No integer $x$ such that $(x-y)^3+ x^3 = (x+y)^3$ 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?
 A: A solution would be a special case of Fermat's last therorem, so there are none.
A: 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...
A: 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.
A: 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.
A: 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., $(x-y)^{23}+x^{23}=(x+y)^{23}$.
A: 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
