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It seems there is no integer $x$ such that such that $(x-y)^3+ x^3 = (x+y)^3$ where $y$ is a non-zero integer. At least I can't find one.

Am I right and if so, how can one show it?

Cameron Williams
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6 Answers6

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We can divide by the cubes of all eventual common factors of $x$ and $y$, so w.l.o.g. $\gcd(x,y)=1$. Expanding all those cubes and rearranging gives $$ x^3=2y^3+6x^2y. $$ The right hand side is always even, so the left hand side must be also. Therefore $x$ is even. But then the left hand side is divisible by $8$ as is the term $6x^2y$. This implies that $2y^3$ must also be divisible by $8$ forcing $y$ to be even, too. So $2\mid \gcd(x,y)$ contradicting our assumption.

Jyrki Lahtonen
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A solution would be a special case of Fermat's last therorem, so there are none.

Peter
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    Is there an elementary proof for this special case? I feel there might be. For example, I think one can show that for any solution both $x$ and $y$ must be even which might be a start. –  Aug 13 '15 at 21:00
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    I only know that the case $n=3$ was solved long before the theorem was solved. If I remember right, already Euler solved this case. – Peter Aug 13 '15 at 21:01
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Rewrite as $$ (x+y)^3 - (x-y)^3 = 2 y (3 x^2 + y^2) = x^3$$ In particular, any prime factor of $y$ divides $x$. But if there is a solution, there is a minimal one...

Robert Israel
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  • Alternatively observe that the original equation (and indeed the derived equation) is homogeneous of degree $3$ - so any common divisor can be cancelled and $x$ and $y$ can be assumed to be coprime. – Mark Bennet Aug 13 '15 at 21:14
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There are only a finite number of possible values of $t = y/x$, just compute them and see they are irrational. No number theory required, and the same method works in fields where Fermat(3) might be false.

$(1-t)^3 + 1 = (1+t)^3$

$2t^3 + 6t^2 - 1 = 0$.

$T^3 + 6T^2 - 4 = 0$ for $T = 2t$.

There are no integer roots.

ASCII Advocate
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    We would want rational roots here, not just integer since you've taken a ratio of $x$ and $y$. However in this case, the rational root theorem says that the only rational roots can be $\pm 1,\pm 2,\pm 4$. – Cameron Williams Aug 13 '15 at 23:54
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    Yes. But we have a monic polynomial so its rational roots must be integers. As specified in the first sentence, we are computing all possible complex values of $t$ and checking that they are irrational. – ASCII Advocate Aug 13 '15 at 23:56
  • Quite right. I was just editing my comment as you said that. I like this answer a lot too. $+1$ – Cameron Williams Aug 13 '15 at 23:57
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Reducing the equation mod $2$ gives

$$x-y+x\equiv x+y\mod2$$

which implies $x\equiv0$ mod $2$. As noted in Jyrki Lahtonen's answer, we can assume $\gcd(x,y)=1$ (i.e., you can remove any common factor in a solution and still have a solution), hence we can assume $y\equiv1$ mod $2$. In particular, $x-y$ and $x+y$ are both odd. But now, reducing things mod $8$ (with the knowledge that $a^3\equiv a$ mod $8$ when $a$ is odd), we have

$$x-y+0\equiv x+y\mod 8$$

which implies $2y\equiv0$ mod $8$, a contradiction to $y\equiv1$ mod $2$.

Remark: This approach works if the cube is replaced with any odd exponent greater than $1$, e.g., $(x-y)^{23}+x^{23}=(x+y)^{23}$.

Barry Cipra
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working it out you get to:

$$x = \left(2+\frac{1}{3}\sqrt[3]{243-27\sqrt{17}}+\sqrt[3]{9+\sqrt{17}}\right)y $$
you will notice that what's inside the brackets is non-integer so the question becomes; is there any integer y for which x is an integer. written another way; can the expression within the brackets be written as a fraction of two integers? i.e is the expression rational

$$\frac{x}{y} = \left(2+\frac{1}{3}\sqrt[3]{243-27\sqrt{17}}+\sqrt[3]{9+\sqrt{17}}\right) $$


which i think boils down to weather:

$$ \sqrt[3]{9+\sqrt{17}}+\sqrt[3]{9-\sqrt{17}} $$


is rational and that i don't know how to prove but ill think about it

peterjtk
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  • It would be much better to typeset your equations using mathjax: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – coldnumber Aug 13 '15 at 23:51