<s>[INST] How many primes are there of the form $x^6+y^6$, where $x,y\in\mathbb{Z}$? I'm trying to find such primes that are of the form $x^6+y^6$; $x,y\in\mathbb{Z}$.
If one of $x$ and $y$ is $0$, say $y=0$ then $x^6+y^6=x^6$ which is not a prime. So assume that $(x,y)\ne(0,0).$
I factored $x^6+y^6$ as follows:
$x^6+y^6=(x^2)^3+(y^2)^3=(x^2+y^2)(x^4+y^4-x^2y^2)=(x^2+y^2)[(x^2+y^2)^2-3x^2y^2]=(x^2+y^2)(x^2+y^2+\sqrt{3}x^2y^2)(x^2+y^2-\sqrt{3}x^2y^2)$
If $x^4+y^4-x^2y^2=1$ then $(x^2+y^2+\sqrt{3}x^2y^2)(x^2+y^2-\sqrt{3}x^2y^2)=1$ this implies that the two numbers $x^2+y^2+\sqrt{3}x^2y^2$ and $x^2+y^2-\sqrt{3}x^2y^2$ are reciprocal to each other. So if $x^2+y^2+\sqrt{3}x^2y^2\in\mathbb{Z}$ then $x^2+y^2-\sqrt{3}x^2y^2\not\in\mathbb{Z}$ or vise-versa.
But $x^2+y^2\pm\sqrt{3}x^2y^2\not\in\mathbb{Z}$ for $(x,y)\ne(0,0)$.
So that $x^4+y^4-x^2y^2$ can't be factored out in $\mathbb{Z}$. That is we can conclude that this is a prime.
To $x^6+y^6$ be a prime, we must have $x^2+y^2=1$, which is possible only for $(x,y)=(+1,0);(-1,0);(0,+1);(0,-1)$.
But in all the above cases, $x^6+y^6=1$, which is not a prime.
So I can conclude that there is no prime of the form $x^6+y^6.$
My question is my argument ok? And is there any other way to think about this problem?
 [/INST] If $x=0$ or $y=0$ then $x^6+y^6$ is $x^6$ or $y^6$ which cannot be a prime.
If $x\ne 0\ne y$ and $x^6+y^6$ is prime then $$x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)=(x^2+y^2)(\,(x^2-y^2)^2+x^2y^2
)$$ and the factor $x^2+y^2\ge 2,$ so the factor $(x^2-y^2)^2+x^2y^2$ must be $1,$ which is not possible unless $|x|=|y|=1$ because $$(x^2-y^2)^2+x^2y^2\ge x^2y^2.$$ So $|x|=|y|=1$ and $x^6+y^6=2.$
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