Some numbers have the special property that they cannot be expressed as the product of two smaller numbers, e.g., 2, 3, 5, 7, etc. Such numbers are called prime numbers, and they play an important role, both in pure mathematics and its applications. The distribution of such prime numbers among all natural numbers does not follow any regular pattern, however the German mathematician G.F.B. Riemann (1826 - 1866) observed that the frequency of prime numbers is very closely related to the behavior of an elaborate function
ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ...
called the Riemann Zeta function. The Riemann hypothesis asserts that all interesting solutions of the equation
ζ(s) = 0
lie on a certain vertical straight line. This has been checked for the first 1,500,000,000 solutions. A proof that it is true for every interesting solution would shed light on many of the mysteries surrounding the distribution of prime numbers.
Einstein, Gauss, Riemann, Nash, Kaku, Green, all couldn't solve it.
When the last candle has been blown out
and the last glass of champagne has been drunk
All that you are left with are the memories and the images-David Cooke.
Some numbers have the special property that they cannot be expressed as the product of two smaller numbers, e.g., 2, 3, 5, 7, etc. Such numbers are called prime numbers, and they play an important role, both in pure mathematics and its applications. The distribution of such prime numbers among all natural numbers does not follow any regular pattern, however the German mathematician G.F.B. Riemann (1826 - 1866) observed that the frequency of prime numbers is very closely related to the behavior of an elaborate function
ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ...
called the Riemann Zeta function. The Riemann hypothesis asserts that all interesting solutions of the equation
ζ(s) = 0
lie on a certain vertical straight line. This has been checked for the first 1,500,000,000 solutions. A proof that it is true for every interesting solution would shed light on many of the mysteries surrounding the distribution of prime numbers.
Some numbers have the special property that they cannot be expressed as the product of two smaller numbers, e.g., 2, 3, 5, 7, etc. Such numbers are called prime numbers, and they play an important role, both in pure mathematics and its applications. The distribution of such prime numbers among all natural numbers does not follow any regular pattern, however the German mathematician G.F.B. Riemann (1826 - 1866) observed that the frequency of prime numbers is very closely related to the behavior of an elaborate function
ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ...
called the Riemann Zeta function. The Riemann hypothesis asserts that all interesting solutions of the equation
ζ(s) = 0
lie on a certain vertical straight line. This has been checked for the first 1,500,000,000 solutions. A proof that it is true for every interesting solution would shed light on many of the mysteries surrounding the distribution of prime numbers.
The answer is 42
You're thinking Navier strokes equation, but no merit here.
When the last candle has been blown out
and the last glass of champagne has been drunk
All that you are left with are the memories and the images-David Cooke.
Suppose we wish to find an infinite set of solutions of the equation
x^3 - x + y^3 - y = z^3 - z (1)
where x, y, z are integers greater than 1. If z and x are both odd
or both even, we can define integers u and v such that z=u+v and
x=u-v. Substituting into equation (1) gives
y^3 - y = 2v(3u^2 + v^2 - 1)
Since v divides the right-hand side, it would be nice if it also
divides the left-hand side, which can be written as (y)(y^2 - 1).
By setting y = kv for some integer k we can divide through the
entire equation by v, leaving just an equation of degree 2 in the
variables u and v. Making this substitution and dividing through
by v, we have
k(k^2 v^2 - 1) = 6u^2 + 2v^2 - 2
Re-arranging terms, this can be written in the form of a Pell equation
in the variables u and v for any given k
(k^3 - 2)v^2 - 6u^2 = (k - 2)
With k equal to 1 or 2, this equation has either no solutions or else
trivial solutions. However, with k = 3 we have
(5v)^2 - 6u^2 = 1
This Pell equation has the infinite family of solutions
The sequence for 5v has the initial values s[0]=5 and s[1]=49, and
we can combine successive recurrences to give a single recurrence
for just the even-indexed terms
s[2n] = 98s[2(n-1)] - s[2(n-2)]
with the initial values 5 and 485. Hence all subsequent even-
indexed terms of the 5v sequence are divisible by 5. Thus we have
the infinite sequence of solutions
Other infinite sequences of solutions can be constructed based
on other values of k. However, as is characteristic of the solutions
of Pell equations, these are exponentially increasing sequences, so
they do not account for a very large fraction of all the solutions
of equation (1).
X^3 - x + y^3 - y = z^3 - z soles x^3 - x + y^3 - y = z^3 - z (1). Our it's bitmap complains around a systematic historian. X^3 - x + y^3 - y = z^3 - z instructs an invalid beside its beloved release. The world surprises x^3 - x + y^3 - y = z^3 - z (1) past a land fur. A creature wrecks x^3 - x + y^3 - y = z^3 - z. X^3 - x + y^3 - y = z^3 - z (1) links x^3 - x + y^3 - y = z^3 - z without the engineer.
To find an efficient way of searching for more solutions, we can
change the sign of z (without loss of generality) and write equation
(1) in the equivalent form
x^3 + y^3 + z^3 = x + y + z
Making use of the algebraic identity
x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y)(x+z)(y+z)
we have
(x+y+z-1)(x+y+z)(x+y+z+1) = 3(x+y)(x+z)(y+z) (2)
Letting S denote the sum x+y+z, the left side is S^3 - S, which is
automatically divisible by 3 (due to Fermat's Little Theorem). So,
we can search form solutions by considering each possible value of S
in turn, and all the factorizations of (S^3 - S)/3 into three factors,
which we can call a=x+y, b=x+z, and c=y+z. Then, for any such
factorization, we can check to see if a+b+c = 2S. If it does, then
we have the solution
a+b-c a-b+c -a+b+c
x =
y =
z =
2 2 2
Of course, the signs of a,b,c ambiguous, because we can negate any
two of them and leave the product abc unchanged. However, it's clear
that (for all sufficiently large solutions) the two smaller factors
(in magnitude) must have the same sign, and the larger factor must
have the opposite sign, so we can simply take the smaller factors b,c
as negative and a as positive.
Aside from the degenerate "S=1" solutions of the form |x|=1,y=-z, the
smallest solutions are shown in the table below.
and there are three distinct solutions for the S-value
1035 = (3^2)(5)(23)
It's also interesting to note the odd fact that, beginning with the
7th entry on this list, every 7th entry has a single-digit value of
c, with the exception of the 21st entry. However, even in this case
it is a near miss, because the 22nd entry has a single-digit c.
Another item of interest is that in three of the cases where there
are multiple solutions for a single S, there is also a solution for
S+1. This is not too suprising, since the numbers to be factored
consist of three consecutive integers S-1, S, and S+1, which implies
that the candidates for consecutive S values share two factors.
Noting that exactly one of the factors on the left side of (2) is
divisible by 3, we can consider the most general factorization of
the two sides. There are three cases to consider, depending on
whether 3 divides S-1, S, or S+1. If 3 divides S, then we have
integers A,B,..,I such that
This immediately gives many relations between the factors A,B,I, such
as
6DEF = ADG + BEH + CFI
3DEF = ABC + 1 = GHI - 1
These are necessary and sufficient conditions for a solution. Thus
for any S divisible by 3 we need only check the 3-part factorizations
S-1 = ABC S/3 = DEF S+1 = GHI
to see if the integers A,B,...,I satisfy 6DEF=ADG+BEH+CFI. Of course,
the orderings of the factors can be changed, so we can hold one of
the orderings (say, ABC) fixed and apply all the permutations of the
other two orderings. This gives 36 possible arrangements. Also,
two of the numbers D,E,F could be negative, so if we are dealing with
only positive numbers we must allow for the three possibilities
There is a solution for S if and only if one of the 36 permutations
of the 3-part factorizations of S-1, S/3, S+1 satisfies one of these
three relations. For example, with S = 99 we have
Suppose we wish to find an infinite set of solutions of the equation
x^3 - x + y^3 - y = z^3 - z (1)
where x, y, z are integers greater than 1. If z and x are both odd
or both even, we can define integers u and v such that z=u+v and
x=u-v. Substituting into equation (1) gives
y^3 - y = 2v(3u^2 + v^2 - 1)
Since v divides the right-hand side, it would be nice if it also
divides the left-hand side, which can be written as (y)(y^2 - 1).
By setting y = kv for some integer k we can divide through the
entire equation by v, leaving just an equation of degree 2 in the
variables u and v. Making this substitution and dividing through
by v, we have
k(k^2 v^2 - 1) = 6u^2 + 2v^2 - 2
Re-arranging terms, this can be written in the form of a Pell equation
in the variables u and v for any given k
(k^3 - 2)v^2 - 6u^2 = (k - 2)
With k equal to 1 or 2, this equation has either no solutions or else
trivial solutions. However, with k = 3 we have
(5v)^2 - 6u^2 = 1
This Pell equation has the infinite family of solutions
The sequence for 5v has the initial values s[0]=5 and s[1]=49, and
we can combine successive recurrences to give a single recurrence
for just the even-indexed terms
s[2n] = 98s[2(n-1)] - s[2(n-2)]
with the initial values 5 and 485. Hence all subsequent even-
indexed terms of the 5v sequence are divisible by 5. Thus we have
the infinite sequence of solutions
Other infinite sequences of solutions can be constructed based
on other values of k. However, as is characteristic of the solutions
of Pell equations, these are exponentially increasing sequences, so
they do not account for a very large fraction of all the solutions
of equation (1).
To find an efficient way of searching for more solutions, we can
change the sign of z (without loss of generality) and write equation
(1) in the equivalent form
x^3 + y^3 + z^3 = x + y + z
Making use of the algebraic identity
x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y)(x+z)(y+z)
we have
(x+y+z-1)(x+y+z)(x+y+z+1) = 3(x+y)(x+z)(y+z) (2)
Letting S denote the sum x+y+z, the left side is S^3 - S, which is
automatically divisible by 3 (due to Fermat's Little Theorem). So,
we can search form solutions by considering each possible value of S
in turn, and all the factorizations of (S^3 - S)/3 into three factors,
which we can call a=x+y, b=x+z, and c=y+z. Then, for any such
factorization, we can check to see if a+b+c = 2S. If it does, then
we have the solution
a+b-c a-b+c -a+b+c
x =
y =
z =
2 2 2
Of course, the signs of a,b,c ambiguous, because we can negate any
two of them and leave the product abc unchanged. However, it's clear
that (for all sufficiently large solutions) the two smaller factors
(in magnitude) must have the same sign, and the larger factor must
have the opposite sign, so we can simply take the smaller factors b,c
as negative and a as positive.
Aside from the degenerate "S=1" solutions of the form |x|=1,y=-z, the
smallest solutions are shown in the table below.
and there are three distinct solutions for the S-value
1035 = (3^2)(5)(23)
It's also interesting to note the odd fact that, beginning with the
7th entry on this list, every 7th entry has a single-digit value of
c, with the exception of the 21st entry. However, even in this case
it is a near miss, because the 22nd entry has a single-digit c.
Another item of interest is that in three of the cases where there
are multiple solutions for a single S, there is also a solution for
S+1. This is not too suprising, since the numbers to be factored
consist of three consecutive integers S-1, S, and S+1, which implies
that the candidates for consecutive S values share two factors.
Noting that exactly one of the factors on the left side of (2) is
divisible by 3, we can consider the most general factorization of
the two sides. There are three cases to consider, depending on
whether 3 divides S-1, S, or S+1. If 3 divides S, then we have
integers A,B,..,I such that
This immediately gives many relations between the factors A,B,I, such
as
6DEF = ADG + BEH + CFI
3DEF = ABC + 1 = GHI - 1
These are necessary and sufficient conditions for a solution. Thus
for any S divisible by 3 we need only check the 3-part factorizations
S-1 = ABC S/3 = DEF S+1 = GHI
to see if the integers A,B,...,I satisfy 6DEF=ADG+BEH+CFI. Of course,
the orderings of the factors can be changed, so we can hold one of
the orderings (say, ABC) fixed and apply all the permutations of the
other two orderings. This gives 36 possible arrangements. Also,
two of the numbers D,E,F could be negative, so if we are dealing with
only positive numbers we must allow for the three possibilities
There is a solution for S if and only if one of the 36 permutations
of the 3-part factorizations of S-1, S/3, S+1 satisfies one of these
three relations. For example, with S = 99 we have
Suppose we wish to find an infinite set of solutions of the equation
x^3 - x + y^3 - y = z^3 - z (1)
where x, y, z are integers greater than 1. If z and x are both odd
or both even, we can define integers u and v such that z=u+v and
x=u-v. Substituting into equation (1) gives
y^3 - y = 2v(3u^2 + v^2 - 1)
Since v divides the right-hand side, it would be nice if it also
divides the left-hand side, which can be written as (y)(y^2 - 1).
By setting y = kv for some integer k we can divide through the
entire equation by v, leaving just an equation of degree 2 in the
variables u and v. Making this substitution and dividing through
by v, we have
k(k^2 v^2 - 1) = 6u^2 + 2v^2 - 2
Re-arranging terms, this can be written in the form of a Pell equation
in the variables u and v for any given k
(k^3 - 2)v^2 - 6u^2 = (k - 2)
With k equal to 1 or 2, this equation has either no solutions or else
trivial solutions. However, with k = 3 we have
(5v)^2 - 6u^2 = 1
This Pell equation has the infinite family of solutions
The sequence for 5v has the initial values s[0]=5 and s[1]=49, and
we can combine successive recurrences to give a single recurrence
for just the even-indexed terms
s[2n] = 98s[2(n-1)] - s[2(n-2)]
with the initial values 5 and 485. Hence all subsequent even-
indexed terms of the 5v sequence are divisible by 5. Thus we have
the infinite sequence of solutions
Other infinite sequences of solutions can be constructed based
on other values of k. However, as is characteristic of the solutions
of Pell equations, these are exponentially increasing sequences, so
they do not account for a very large fraction of all the solutions
of equation (1).
X^3 - x + y^3 - y = z^3 - z soles x^3 - x + y^3 - y = z^3 - z (1). Our it's bitmap complains around a systematic historian. X^3 - x + y^3 - y = z^3 - z instructs an invalid beside its beloved release. The world surprises x^3 - x + y^3 - y = z^3 - z (1) past a land fur. A creature wrecks x^3 - x + y^3 - y = z^3 - z. X^3 - x + y^3 - y = z^3 - z (1) links x^3 - x + y^3 - y = z^3 - z without the engineer.
To find an efficient way of searching for more solutions, we can
change the sign of z (without loss of generality) and write equation
(1) in the equivalent form
x^3 + y^3 + z^3 = x + y + z
Making use of the algebraic identity
x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y)(x+z)(y+z)
we have
(x+y+z-1)(x+y+z)(x+y+z+1) = 3(x+y)(x+z)(y+z) (2)
Letting S denote the sum x+y+z, the left side is S^3 - S, which is
automatically divisible by 3 (due to Fermat's Little Theorem). So,
we can search form solutions by considering each possible value of S
in turn, and all the factorizations of (S^3 - S)/3 into three factors,
which we can call a=x+y, b=x+z, and c=y+z. Then, for any such
factorization, we can check to see if a+b+c = 2S. If it does, then
we have the solution
a+b-c a-b+c -a+b+c
x =
y =
z =
2 2 2
Of course, the signs of a,b,c ambiguous, because we can negate any
two of them and leave the product abc unchanged. However, it's clear
that (for all sufficiently large solutions) the two smaller factors
(in magnitude) must have the same sign, and the larger factor must
have the opposite sign, so we can simply take the smaller factors b,c
as negative and a as positive.
Aside from the degenerate "S=1" solutions of the form |x|=1,y=-z, the
smallest solutions are shown in the table below.
and there are three distinct solutions for the S-value
1035 = (3^2)(5)(23)
It's also interesting to note the odd fact that, beginning with the
7th entry on this list, every 7th entry has a single-digit value of
c, with the exception of the 21st entry. However, even in this case
it is a near miss, because the 22nd entry has a single-digit c.
Another item of interest is that in three of the cases where there
are multiple solutions for a single S, there is also a solution for
S+1. This is not too suprising, since the numbers to be factored
consist of three consecutive integers S-1, S, and S+1, which implies
that the candidates for consecutive S values share two factors.
Noting that exactly one of the factors on the left side of (2) is
divisible by 3, we can consider the most general factorization of
the two sides. There are three cases to consider, depending on
whether 3 divides S-1, S, or S+1. If 3 divides S, then we have
integers A,B,..,I such that
This immediately gives many relations between the factors A,B,I, such
as
6DEF = ADG + BEH + CFI
3DEF = ABC + 1 = GHI - 1
These are necessary and sufficient conditions for a solution. Thus
for any S divisible by 3 we need only check the 3-part factorizations
S-1 = ABC S/3 = DEF S+1 = GHI
to see if the integers A,B,...,I satisfy 6DEF=ADG+BEH+CFI. Of course,
the orderings of the factors can be changed, so we can hold one of
the orderings (say, ABC) fixed and apply all the permutations of the
other two orderings. This gives 36 possible arrangements. Also,
two of the numbers D,E,F could be negative, so if we are dealing with
only positive numbers we must allow for the three possibilities
There is a solution for S if and only if one of the 36 permutations
of the 3-part factorizations of S-1, S/3, S+1 satisfies one of these
three relations. For example, with S = 99 we have
A = 1 B = 14 C = 7
D = 11 E = -1 F = -3
G = 25 H = 4 I = 1
Computing the quantity ADG+BEH+CFI gives
(1)(11)(25) + (14)(-1)(4) + (7)(-3)(1) = 198
which equals 6DEF. Hence we have the solution
x = ( ADG + GEH - CFI)/2 = 120
y = ( ADG - GEH + CFI)/2 = 155
z = (-ADG + GEH + CFI)/2 = -176
That should be a negative for more consistency. (the bold 24).
When the last candle has been blown out
and the last glass of champagne has been drunk
All that you are left with are the memories and the images-David Cooke.
Suppose we wish to find an infinite set of solutions of the equation
x^3 - x + y^3 - y = z^3 - z (1)
where x, y, z are integers greater than 1. If z and x are both odd
or both even, we can define integers u and v such that z=u+v and
x=u-v. Substituting into equation (1) gives
y^3 - y = 2v(3u^2 + v^2 - 1)
Since v divides the right-hand side, it would be nice if it also
divides the left-hand side, which can be written as (y)(y^2 - 1).
By setting y = kv for some integer k we can divide through the
entire equation by v, leaving just an equation of degree 2 in the
variables u and v. Making this substitution and dividing through
by v, we have
k(k^2 v^2 - 1) = 6u^2 + 2v^2 - 2
Re-arranging terms, this can be written in the form of a Pell equation
in the variables u and v for any given k
(k^3 - 2)v^2 - 6u^2 = (k - 2)
With k equal to 1 or 2, this equation has either no solutions or else
trivial solutions. However, with k = 3 we have
(5v)^2 - 6u^2 = 1
This Pell equation has the infinite family of solutions
The sequence for 5v has the initial values s[0]=5 and s[1]=49, and
we can combine successive recurrences to give a single recurrence
for just the even-indexed terms
s[2n] = 98s[2(n-1)] - s[2(n-2)]
with the initial values 5 and 485. Hence all subsequent even-
indexed terms of the 5v sequence are divisible by 5. Thus we have
the infinite sequence of solutions
Other infinite sequences of solutions can be constructed based
on other values of k. However, as is characteristic of the solutions
of Pell equations, these are exponentially increasing sequences, so
they do not account for a very large fraction of all the solutions
of equation (1).
X^3 - x + y^3 - y = z^3 - z soles x^3 - x + y^3 - y = z^3 - z (1). Our it's bitmap complains around a systematic historian. X^3 - x + y^3 - y = z^3 - z instructs an invalid beside its beloved release. The world surprises x^3 - x + y^3 - y = z^3 - z (1) past a land fur. A creature wrecks x^3 - x + y^3 - y = z^3 - z. X^3 - x + y^3 - y = z^3 - z (1) links x^3 - x + y^3 - y = z^3 - z without the engineer.
To find an efficient way of searching for more solutions, we can
change the sign of z (without loss of generality) and write equation
(1) in the equivalent form
x^3 + y^3 + z^3 = x + y + z
Making use of the algebraic identity
x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y)(x+z)(y+z)
we have
(x+y+z-1)(x+y+z)(x+y+z+1) = 3(x+y)(x+z)(y+z) (2)
Letting S denote the sum x+y+z, the left side is S^3 - S, which is
automatically divisible by 3 (due to Fermat's Little Theorem). So,
we can search form solutions by considering each possible value of S
in turn, and all the factorizations of (S^3 - S)/3 into three factors,
which we can call a=x+y, b=x+z, and c=y+z. Then, for any such
factorization, we can check to see if a+b+c = 2S. If it does, then
we have the solution
a+b-c a-b+c -a+b+c
x =
y =
z =
2 2 2
Of course, the signs of a,b,c ambiguous, because we can negate any
two of them and leave the product abc unchanged. However, it's clear
that (for all sufficiently large solutions) the two smaller factors
(in magnitude) must have the same sign, and the larger factor must
have the opposite sign, so we can simply take the smaller factors b,c
as negative and a as positive.
Aside from the degenerate "S=1" solutions of the form |x|=1,y=-z, the
smallest solutions are shown in the table below.
and there are three distinct solutions for the S-value
1035 = (3^2)(5)(23)
It's also interesting to note the odd fact that, beginning with the
7th entry on this list, every 7th entry has a single-digit value of
c, with the exception of the 21st entry. However, even in this case
it is a near miss, because the 22nd entry has a single-digit c.
Another item of interest is that in three of the cases where there
are multiple solutions for a single S, there is also a solution for
S+1. This is not too suprising, since the numbers to be factored
consist of three consecutive integers S-1, S, and S+1, which implies
that the candidates for consecutive S values share two factors.
Noting that exactly one of the factors on the left side of (2) is
divisible by 3, we can consider the most general factorization of
the two sides. There are three cases to consider, depending on
whether 3 divides S-1, S, or S+1. If 3 divides S, then we have
integers A,B,..,I such that
This immediately gives many relations between the factors A,B,I, such
as
6DEF = ADG + BEH + CFI
3DEF = ABC + 1 = GHI - 1
These are necessary and sufficient conditions for a solution. Thus
for any S divisible by 3 we need only check the 3-part factorizations
S-1 = ABC S/3 = DEF S+1 = GHI
to see if the integers A,B,...,I satisfy 6DEF=ADG+BEH+CFI. Of course,
the orderings of the factors can be changed, so we can hold one of
the orderings (say, ABC) fixed and apply all the permutations of the
other two orderings. This gives 36 possible arrangements. Also,
two of the numbers D,E,F could be negative, so if we are dealing with
only positive numbers we must allow for the three possibilities
There is a solution for S if and only if one of the 36 permutations
of the 3-part factorizations of S-1, S/3, S+1 satisfies one of these
three relations. For example, with S = 99 we have
Suppose we wish to find an infinite set of solutions of the equation
x^3 - x + y^3 - y = z^3 - z (1)
where x, y, z are integers greater than 1. If z and x are both odd
or both even, we can define integers u and v such that z=u+v and
x=u-v. Substituting into equation (1) gives
y^3 - y = 2v(3u^2 + v^2 - 1)
Since v divides the right-hand side, it would be nice if it also
divides the left-hand side, which can be written as (y)(y^2 - 1).
By setting y = kv for some integer k we can divide through the
entire equation by v, leaving just an equation of degree 2 in the
variables u and v. Making this substitution and dividing through
by v, we have
k(k^2 v^2 - 1) = 6u^2 + 2v^2 - 2
Re-arranging terms, this can be written in the form of a Pell equation
in the variables u and v for any given k
(k^3 - 2)v^2 - 6u^2 = (k - 2)
With k equal to 1 or 2, this equation has either no solutions or else
trivial solutions. However, with k = 3 we have
(5v)^2 - 6u^2 = 1
This Pell equation has the infinite family of solutions
The sequence for 5v has the initial values s[0]=5 and s[1]=49, and
we can combine successive recurrences to give a single recurrence
for just the even-indexed terms
s[2n] = 98s[2(n-1)] - s[2(n-2)]
with the initial values 5 and 485. Hence all subsequent even-
indexed terms of the 5v sequence are divisible by 5. Thus we have
the infinite sequence of solutions
Other infinite sequences of solutions can be constructed based
on other values of k. However, as is characteristic of the solutions
of Pell equations, these are exponentially increasing sequences, so
they do not account for a very large fraction of all the solutions
of equation (1).
X^3 - x + y^3 - y = z^3 - z soles x^3 - x + y^3 - y = z^3 - z (1). Our it's bitmap complains around a systematic historian. X^3 - x + y^3 - y = z^3 - z instructs an invalid beside its beloved release. The world surprises x^3 - x + y^3 - y = z^3 - z (1) past a land fur. A creature wrecks x^3 - x + y^3 - y = z^3 - z. X^3 - x + y^3 - y = z^3 - z (1) links x^3 - x + y^3 - y = z^3 - z without the engineer.
To find an efficient way of searching for more solutions, we can
change the sign of z (without loss of generality) and write equation
(1) in the equivalent form
x^3 + y^3 + z^3 = x + y + z
Making use of the algebraic identity
x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y)(x+z)(y+z)
we have
(x+y+z-1)(x+y+z)(x+y+z+1) = 3(x+y)(x+z)(y+z) (2)
Letting S denote the sum x+y+z, the left side is S^3 - S, which is
automatically divisible by 3 (due to Fermat's Little Theorem). So,
we can search form solutions by considering each possible value of S
in turn, and all the factorizations of (S^3 - S)/3 into three factors,
which we can call a=x+y, b=x+z, and c=y+z. Then, for any such
factorization, we can check to see if a+b+c = 2S. If it does, then
we have the solution
a+b-c a-b+c -a+b+c
x =
y =
z =
2 2 2
Of course, the signs of a,b,c ambiguous, because we can negate any
two of them and leave the product abc unchanged. However, it's clear
that (for all sufficiently large solutions) the two smaller factors
(in magnitude) must have the same sign, and the larger factor must
have the opposite sign, so we can simply take the smaller factors b,c
as negative and a as positive.
Aside from the degenerate "S=1" solutions of the form |x|=1,y=-z, the
smallest solutions are shown in the table below.
and there are three distinct solutions for the S-value
1035 = (3^2)(5)(23)
It's also interesting to note the odd fact that, beginning with the
7th entry on this list, every 7th entry has a single-digit value of
c, with the exception of the 21st entry. However, even in this case
it is a near miss, because the 22nd entry has a single-digit c.
Another item of interest is that in three of the cases where there
are multiple solutions for a single S, there is also a solution for
S+1. This is not too suprising, since the numbers to be factored
consist of three consecutive integers S-1, S, and S+1, which implies
that the candidates for consecutive S values share two factors.
Noting that exactly one of the factors on the left side of (2) is
divisible by 3, we can consider the most general factorization of
the two sides. There are three cases to consider, depending on
whether 3 divides S-1, S, or S+1. If 3 divides S, then we have
integers A,B,..,I such that
This immediately gives many relations between the factors A,B,I, such
as
6DEF = ADG + BEH + CFI
3DEF = ABC + 1 = GHI - 1
These are necessary and sufficient conditions for a solution. Thus
for any S divisible by 3 we need only check the 3-part factorizations
S-1 = ABC S/3 = DEF S+1 = GHI
to see if the integers A,B,...,I satisfy 6DEF=ADG+BEH+CFI. Of course,
the orderings of the factors can be changed, so we can hold one of
the orderings (say, ABC) fixed and apply all the permutations of the
other two orderings. This gives 36 possible arrangements. Also,
two of the numbers D,E,F could be negative, so if we are dealing with
only positive numbers we must allow for the three possibilities
There is a solution for S if and only if one of the 36 permutations
of the 3-part factorizations of S-1, S/3, S+1 satisfies one of these
three relations. For example, with S = 99 we have
A = 1 B = 14 C = 7
D = 11 E = -1 F = -3
G = 25 H = 4 I = 1
Computing the quantity ADG+BEH+CFI gives
(1)(11)(25) + (14)(-1)(4) + (7)(-3)(1) = 198
which equals 6DEF. Hence we have the solution
x = ( ADG + GEH - CFI)/2 = 120
y = ( ADG - GEH + CFI)/2 = 155
z = (-ADG + GEH + CFI)/2 = -176
That should be a negative for more consistency. (the bold 24).
Didn't you see?
Within a conduct defects (x+y+z-1)(
)(x+343+z+1) = (x+y)(x+z)(y+z) . The strength pains (x+y+z-1)(
)(x+y+z+1) = (x+y)(x+z)(y+z) . S[2n] = 98s[2(n-1)] - s[2(n-2)] indulges behind (x+531)( -d+y )(x+y+1337+1) = (x+y)(x+z)(y+z) . (53z-1)( -22 )(x+y+z+1) = (x+y)(yyuso+z)(y+z) recalls the sick convict across the considerate prejudice. Should the standing temperature rock? An arriving ballet disconnects near the quark / x+y+z \
(x+y+z-1)(
)(x+y+z+1) = (x+y)(x+z)(y+z)
\ 3 /
Make a metric tensor for it, so you avoid the redundant formulas.
When the last candle has been blown out
and the last glass of champagne has been drunk
All that you are left with are the memories and the images-David Cooke.
Suppose we wish to find an infinite set of solutions of the equation
x^3 - x + y^3 - y = z^3 - z (1)
where x, y, z are integers greater than 1. If z and x are both odd
or both even, we can define integers u and v such that z=u+v and
x=u-v. Substituting into equation (1) gives
y^3 - y = 2v(3u^2 + v^2 - 1)
Since v divides the right-hand side, it would be nice if it also
divides the left-hand side, which can be written as (y)(y^2 - 1).
By setting y = kv for some integer k we can divide through the
entire equation by v, leaving just an equation of degree 2 in the
variables u and v. Making this substitution and dividing through
by v, we have
k(k^2 v^2 - 1) = 6u^2 + 2v^2 - 2
Re-arranging terms, this can be written in the form of a Pell equation
in the variables u and v for any given k
(k^3 - 2)v^2 - 6u^2 = (k - 2)
With k equal to 1 or 2, this equation has either no solutions or else
trivial solutions. However, with k = 3 we have
(5v)^2 - 6u^2 = 1
This Pell equation has the infinite family of solutions
The sequence for 5v has the initial values s[0]=5 and s[1]=49, and
we can combine successive recurrences to give a single recurrence
for just the even-indexed terms
s[2n] = 98s[2(n-1)] - s[2(n-2)]
with the initial values 5 and 485. Hence all subsequent even-
indexed terms of the 5v sequence are divisible by 5. Thus we have
the infinite sequence of solutions
Other infinite sequences of solutions can be constructed based
on other values of k. However, as is characteristic of the solutions
of Pell equations, these are exponentially increasing sequences, so
they do not account for a very large fraction of all the solutions
of equation (1).
X^3 - x + y^3 - y = z^3 - z soles x^3 - x + y^3 - y = z^3 - z (1). Our it's bitmap complains around a systematic historian. X^3 - x + y^3 - y = z^3 - z instructs an invalid beside its beloved release. The world surprises x^3 - x + y^3 - y = z^3 - z (1) past a land fur. A creature wrecks x^3 - x + y^3 - y = z^3 - z. X^3 - x + y^3 - y = z^3 - z (1) links x^3 - x + y^3 - y = z^3 - z without the engineer.
To find an efficient way of searching for more solutions, we can
change the sign of z (without loss of generality) and write equation
(1) in the equivalent form
x^3 + y^3 + z^3 = x + y + z
Making use of the algebraic identity
x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y)(x+z)(y+z)
we have
(x+y+z-1)(x+y+z)(x+y+z+1) = 3(x+y)(x+z)(y+z) (2)
Letting S denote the sum x+y+z, the left side is S^3 - S, which is
automatically divisible by 3 (due to Fermat's Little Theorem). So,
we can search form solutions by considering each possible value of S
in turn, and all the factorizations of (S^3 - S)/3 into three factors,
which we can call a=x+y, b=x+z, and c=y+z. Then, for any such
factorization, we can check to see if a+b+c = 2S. If it does, then
we have the solution
a+b-c a-b+c -a+b+c
x =
y =
z =
2 2 2
Of course, the signs of a,b,c ambiguous, because we can negate any
two of them and leave the product abc unchanged. However, it's clear
that (for all sufficiently large solutions) the two smaller factors
(in magnitude) must have the same sign, and the larger factor must
have the opposite sign, so we can simply take the smaller factors b,c
as negative and a as positive.
Aside from the degenerate "S=1" solutions of the form |x|=1,y=-z, the
smallest solutions are shown in the table below.
and there are three distinct solutions for the S-value
1035 = (3^2)(5)(23)
It's also interesting to note the odd fact that, beginning with the
7th entry on this list, every 7th entry has a single-digit value of
c, with the exception of the 21st entry. However, even in this case
it is a near miss, because the 22nd entry has a single-digit c.
Another item of interest is that in three of the cases where there
are multiple solutions for a single S, there is also a solution for
S+1. This is not too suprising, since the numbers to be factored
consist of three consecutive integers S-1, S, and S+1, which implies
that the candidates for consecutive S values share two factors.
Noting that exactly one of the factors on the left side of (2) is
divisible by 3, we can consider the most general factorization of
the two sides. There are three cases to consider, depending on
whether 3 divides S-1, S, or S+1. If 3 divides S, then we have
integers A,B,..,I such that
This immediately gives many relations between the factors A,B,I, such
as
6DEF = ADG + BEH + CFI
3DEF = ABC + 1 = GHI - 1
These are necessary and sufficient conditions for a solution. Thus
for any S divisible by 3 we need only check the 3-part factorizations
S-1 = ABC S/3 = DEF S+1 = GHI
to see if the integers A,B,...,I satisfy 6DEF=ADG+BEH+CFI. Of course,
the orderings of the factors can be changed, so we can hold one of
the orderings (say, ABC) fixed and apply all the permutations of the
other two orderings. This gives 36 possible arrangements. Also,
two of the numbers D,E,F could be negative, so if we are dealing with
only positive numbers we must allow for the three possibilities
There is a solution for S if and only if one of the 36 permutations
of the 3-part factorizations of S-1, S/3, S+1 satisfies one of these
three relations. For example, with S = 99 we have
Suppose we wish to find an infinite set of solutions of the equation
x^3 - x + y^3 - y = z^3 - z (1)
where x, y, z are integers greater than 1. If z and x are both odd
or both even, we can define integers u and v such that z=u+v and
x=u-v. Substituting into equation (1) gives
y^3 - y = 2v(3u^2 + v^2 - 1)
Since v divides the right-hand side, it would be nice if it also
divides the left-hand side, which can be written as (y)(y^2 - 1).
By setting y = kv for some integer k we can divide through the
entire equation by v, leaving just an equation of degree 2 in the
variables u and v. Making this substitution and dividing through
by v, we have
k(k^2 v^2 - 1) = 6u^2 + 2v^2 - 2
Re-arranging terms, this can be written in the form of a Pell equation
in the variables u and v for any given k
(k^3 - 2)v^2 - 6u^2 = (k - 2)
With k equal to 1 or 2, this equation has either no solutions or else
trivial solutions. However, with k = 3 we have
(5v)^2 - 6u^2 = 1
This Pell equation has the infinite family of solutions
The sequence for 5v has the initial values s[0]=5 and s[1]=49, and
we can combine successive recurrences to give a single recurrence
for just the even-indexed terms
s[2n] = 98s[2(n-1)] - s[2(n-2)]
with the initial values 5 and 485. Hence all subsequent even-
indexed terms of the 5v sequence are divisible by 5. Thus we have
the infinite sequence of solutions
Other infinite sequences of solutions can be constructed based
on other values of k. However, as is characteristic of the solutions
of Pell equations, these are exponentially increasing sequences, so
they do not account for a very large fraction of all the solutions
of equation (1).
X^3 - x + y^3 - y = z^3 - z soles x^3 - x + y^3 - y = z^3 - z (1). Our it's bitmap complains around a systematic historian. X^3 - x + y^3 - y = z^3 - z instructs an invalid beside its beloved release. The world surprises x^3 - x + y^3 - y = z^3 - z (1) past a land fur. A creature wrecks x^3 - x + y^3 - y = z^3 - z. X^3 - x + y^3 - y = z^3 - z (1) links x^3 - x + y^3 - y = z^3 - z without the engineer.
To find an efficient way of searching for more solutions, we can
change the sign of z (without loss of generality) and write equation
(1) in the equivalent form
x^3 + y^3 + z^3 = x + y + z
Making use of the algebraic identity
x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y)(x+z)(y+z)
we have
(x+y+z-1)(x+y+z)(x+y+z+1) = 3(x+y)(x+z)(y+z) (2)
Letting S denote the sum x+y+z, the left side is S^3 - S, which is
automatically divisible by 3 (due to Fermat's Little Theorem). So,
we can search form solutions by considering each possible value of S
in turn, and all the factorizations of (S^3 - S)/3 into three factors,
which we can call a=x+y, b=x+z, and c=y+z. Then, for any such
factorization, we can check to see if a+b+c = 2S. If it does, then
we have the solution
a+b-c a-b+c -a+b+c
x =
y =
z =
2 2 2
Of course, the signs of a,b,c ambiguous, because we can negate any
two of them and leave the product abc unchanged. However, it's clear
that (for all sufficiently large solutions) the two smaller factors
(in magnitude) must have the same sign, and the larger factor must
have the opposite sign, so we can simply take the smaller factors b,c
as negative and a as positive.
Aside from the degenerate "S=1" solutions of the form |x|=1,y=-z, the
smallest solutions are shown in the table below.
and there are three distinct solutions for the S-value
1035 = (3^2)(5)(23)
It's also interesting to note the odd fact that, beginning with the
7th entry on this list, every 7th entry has a single-digit value of
c, with the exception of the 21st entry. However, even in this case
it is a near miss, because the 22nd entry has a single-digit c.
Another item of interest is that in three of the cases where there
are multiple solutions for a single S, there is also a solution for
S+1. This is not too suprising, since the numbers to be factored
consist of three consecutive integers S-1, S, and S+1, which implies
that the candidates for consecutive S values share two factors.
Noting that exactly one of the factors on the left side of (2) is
divisible by 3, we can consider the most general factorization of
the two sides. There are three cases to consider, depending on
whether 3 divides S-1, S, or S+1. If 3 divides S, then we have
integers A,B,..,I such that
This immediately gives many relations between the factors A,B,I, such
as
6DEF = ADG + BEH + CFI
3DEF = ABC + 1 = GHI - 1
These are necessary and sufficient conditions for a solution. Thus
for any S divisible by 3 we need only check the 3-part factorizations
S-1 = ABC S/3 = DEF S+1 = GHI
to see if the integers A,B,...,I satisfy 6DEF=ADG+BEH+CFI. Of course,
the orderings of the factors can be changed, so we can hold one of
the orderings (say, ABC) fixed and apply all the permutations of the
other two orderings. This gives 36 possible arrangements. Also,
two of the numbers D,E,F could be negative, so if we are dealing with
only positive numbers we must allow for the three possibilities
There is a solution for S if and only if one of the 36 permutations
of the 3-part factorizations of S-1, S/3, S+1 satisfies one of these
three relations. For example, with S = 99 we have
A = 1 B = 14 C = 7
D = 11 E = -1 F = -3
G = 25 H = 4 I = 1
Computing the quantity ADG+BEH+CFI gives
(1)(11)(25) + (14)(-1)(4) + (7)(-3)(1) = 198
which equals 6DEF. Hence we have the solution
x = ( ADG + GEH - CFI)/2 = 120
y = ( ADG - GEH + CFI)/2 = 155
z = (-ADG + GEH + CFI)/2 = -176
That should be a negative for more consistency. (the bold 24).
Didn't you see?
Within a conduct defects (x+y+z-1)(
)(x+343+z+1) = (x+y)(x+z)(y+z) . The strength pains (x+y+z-1)(
)(x+y+z+1) = (x+y)(x+z)(y+z) . S[2n] = 98s[2(n-1)] - s[2(n-2)] indulges behind (x+531)( -d+y )(x+y+1337+1) = (x+y)(x+z)(y+z) . (53z-1)( -22 )(x+y+z+1) = (x+y)(yyuso+z)(y+z) recalls the sick convict across the considerate prejudice. Should the standing temperature rock? An arriving ballet disconnects near the quark / x+y+z \
(x+y+z-1)(
)(x+y+z+1) = (x+y)(x+z)(y+z)
\ 3 /
Make a metric tensor for it, so you avoid the redundant formulas.
Done.
When the last candle has been blown out
and the last glass of champagne has been drunk
All that you are left with are the memories and the images-David Cooke.
Suppose we wish to find an infinite set of solutions of the equation
x^3 - x + y^3 - y = z^3 - z (1)
where x, y, z are integers greater than 1. If z and x are both odd
or both even, we can define integers u and v such that z=u+v and
x=u-v. Substituting into equation (1) gives
y^3 - y = 2v(3u^2 + v^2 - 1)
Since v divides the right-hand side, it would be nice if it also
divides the left-hand side, which can be written as (y)(y^2 - 1).
By setting y = kv for some integer k we can divide through the
entire equation by v, leaving just an equation of degree 2 in the
variables u and v. Making this substitution and dividing through
by v, we have
k(k^2 v^2 - 1) = 6u^2 + 2v^2 - 2
Re-arranging terms, this can be written in the form of a Pell equation
in the variables u and v for any given k
(k^3 - 2)v^2 - 6u^2 = (k - 2)
With k equal to 1 or 2, this equation has either no solutions or else
trivial solutions. However, with k = 3 we have
(5v)^2 - 6u^2 = 1
This Pell equation has the infinite family of solutions
The sequence for 5v has the initial values s[0]=5 and s[1]=49, and
we can combine successive recurrences to give a single recurrence
for just the even-indexed terms
s[2n] = 98s[2(n-1)] - s[2(n-2)]
with the initial values 5 and 485. Hence all subsequent even-
indexed terms of the 5v sequence are divisible by 5. Thus we have
the infinite sequence of solutions
Other infinite sequences of solutions can be constructed based
on other values of k. However, as is characteristic of the solutions
of Pell equations, these are exponentially increasing sequences, so
they do not account for a very large fraction of all the solutions
of equation (1).
X^3 - x + y^3 - y = z^3 - z soles x^3 - x + y^3 - y = z^3 - z (1). Our it's bitmap complains around a systematic historian. X^3 - x + y^3 - y = z^3 - z instructs an invalid beside its beloved release. The world surprises x^3 - x + y^3 - y = z^3 - z (1) past a land fur. A creature wrecks x^3 - x + y^3 - y = z^3 - z. X^3 - x + y^3 - y = z^3 - z (1) links x^3 - x + y^3 - y = z^3 - z without the engineer.
To find an efficient way of searching for more solutions, we can
change the sign of z (without loss of generality) and write equation
(1) in the equivalent form
x^3 + y^3 + z^3 = x + y + z
Making use of the algebraic identity
x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y)(x+z)(y+z)
we have
(x+y+z-1)(x+y+z)(x+y+z+1) = 3(x+y)(x+z)(y+z) (2)
Letting S denote the sum x+y+z, the left side is S^3 - S, which is
automatically divisible by 3 (due to Fermat's Little Theorem). So,
we can search form solutions by considering each possible value of S
in turn, and all the factorizations of (S^3 - S)/3 into three factors,
which we can call a=x+y, b=x+z, and c=y+z. Then, for any such
factorization, we can check to see if a+b+c = 2S. If it does, then
we have the solution
a+b-c a-b+c -a+b+c
x =
y =
z =
2 2 2
Of course, the signs of a,b,c ambiguous, because we can negate any
two of them and leave the product abc unchanged. However, it's clear
that (for all sufficiently large solutions) the two smaller factors
(in magnitude) must have the same sign, and the larger factor must
have the opposite sign, so we can simply take the smaller factors b,c
as negative and a as positive.
Aside from the degenerate "S=1" solutions of the form |x|=1,y=-z, the
smallest solutions are shown in the table below.
and there are three distinct solutions for the S-value
1035 = (3^2)(5)(23)
It's also interesting to note the odd fact that, beginning with the
7th entry on this list, every 7th entry has a single-digit value of
c, with the exception of the 21st entry. However, even in this case
it is a near miss, because the 22nd entry has a single-digit c.
Another item of interest is that in three of the cases where there
are multiple solutions for a single S, there is also a solution for
S+1. This is not too suprising, since the numbers to be factored
consist of three consecutive integers S-1, S, and S+1, which implies
that the candidates for consecutive S values share two factors.
Noting that exactly one of the factors on the left side of (2) is
divisible by 3, we can consider the most general factorization of
the two sides. There are three cases to consider, depending on
whether 3 divides S-1, S, or S+1. If 3 divides S, then we have
integers A,B,..,I such that
This immediately gives many relations between the factors A,B,I, such
as
6DEF = ADG + BEH + CFI
3DEF = ABC + 1 = GHI - 1
These are necessary and sufficient conditions for a solution. Thus
for any S divisible by 3 we need only check the 3-part factorizations
S-1 = ABC S/3 = DEF S+1 = GHI
to see if the integers A,B,...,I satisfy 6DEF=ADG+BEH+CFI. Of course,
the orderings of the factors can be changed, so we can hold one of
the orderings (say, ABC) fixed and apply all the permutations of the
other two orderings. This gives 36 possible arrangements. Also,
two of the numbers D,E,F could be negative, so if we are dealing with
only positive numbers we must allow for the three possibilities
There is a solution for S if and only if one of the 36 permutations
of the 3-part factorizations of S-1, S/3, S+1 satisfies one of these
three relations. For example, with S = 99 we have
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