<s>[INST] Problems & Solutions on Fermat Theorem of Multiple of 3 I am working on an assignment in elementary number theory, in which I have to come up with original problems and then work out their solutions on Fermat theorem of multiple of 3, that is, the equation
$$x^3 + y^3 = z^3$$
does not have integer solution for $xyz \neq 0$.
The first problem that I came up easily is one requiring readers to prove that $x^6 + y^6 = z^6$ does not have integer solution, which is very straightforward. But I need more than just one problem. I have searched around in the internet for ideas but could not find any. Therefore here is my question:
Do you have any ideas, suggestion, links or hints that perhaps can help me with writing couple more question related to the subject? Thank you for your time and help.
 [/INST] Solve in integers $\left(a-b\right)\left(a^2-b^2\right)\left(a^3-b^3\right)=3c^3$.
To solve it, notice $(3c)^3=(a^2-2b^2+ab)^3 + (2a^2-b^2-ab)^3$, 
so either $c=0$ or $a^2-2b^2+ab=0$ or $2a^2-b^2-ab=0$.
Here's where I found the problem.

$$(a-b)\left(a^2-b^2\right)\left(a^3-b^3\right)=(a-b)^3(a+b)\left(a^2+ab+b^2\right)$$
$a^2+ab+b^2=\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\ge 0$ with equality if and only if $a=b=0$.
If $c=0$, then either $a=b$ or $a=-b$, so $(a,b,c)=(k,k,0),(k,-k,0)$, $k\in\mathbb Z$.
If $a^2-2b^2+ab=0$, then $(2a+b)^2=9b^2$, so $2a+b=\pm 3b$. 
If $2a+b=3b$, then $a=b$, so $(a,b,c)=(k,k,0)$.
If $2a+b=-3b$, then $a=-2b$, so $(a,b,c)=\left(-2k,k,3k^2\right)$.
If $2a^2-b^2-ab=0$, then $b^2-2a^2+ab=0$ (symmetrical to the other case), so analogously $(a,b,c)=(k,k,0),\left(k,-2k,-3k^2\right)$.
All solutions are $(a,b,c)=(k,k,0),(k,-k,0),(-2k,k,3k^2),(k,-2k,-3k^2)$.
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