How many primes are there of the form $x^6+y^6$, where $x,y\in\mathbb{Z}$?

Question

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?

Answer 1

So you know that $x^6 + y^6 = (x^2 + y^2)(x^4 - x^2y^2 + y^4)$, and both are not factorizable over $\mathbb Q$.

If $a, b, c \in \mathbb Z$ and $a = bc$, then if $a$ is prime, either $b = 1$ or $c = 1$.

You already know that $x^2 + y^2 = 1$ if and only if $(x, y) \in \{(0, \pm 1), (\pm 1, 0)\}$. Let us briefly look at the other term.

Actually, $x^4 - x^2y^2 + y^4$ will not be small too often, since $x^4 - x^2y^2 + y^4 = (x^2-y^2)^2 + x^2y^2 \geq x^2y^2$. In particular, it is $1$ only if $(x, y) = (\pm1, \pm 1)$.

Combining all these, the only prime $x^6+y^6$ hits is $2$.

For your other question in your comment, actually one can prove more:

For any multivariate non-constant polynomial $f \in \mathbb Z[x_1, \cdots, x_n]$ with integral coefficients, there exists $(a_1, \cdots, a_n) \in \mathbb Z^n$ such that $(a_i)$ are pairwise relatively prime and $|f(a_1, \cdots, a_n)|$ is composite.

If you care about why, here is a sketch of the construction. The construction can roughly be taken in the following steps:

1. Fix temporarily $a_2 = \cdots = a_n = 1$, and consider the univariate polynomial $f_1(x_1) = f(x_1, 1, \cdots, 1)$. There exists a $a_1 \neq 0$ such that $|f_1(a_1)| \geq 2$. If it is composite, we are done. If not, let $p$ be a prime divisor of $|f_1(a_1)|$.

2. The iteration: For $i = 2, \cdots n$, determine the final value $a_i$ satisfying the following three conditions:

   (a) $a_i \equiv 1 \pmod p$.

   (b) $a_i$ is relatively prime to $a_1, \cdots, a_{i-1}$.

   (c) $|f(a_1, \cdots, a_i, 1, \cdots, 1)| > p$

This is always possible. For the correctness, (a) guarantees us that $p \ | \ |f(a_1, \cdots, a_i, 1, \cdots, 1)|$; (b) guarantees everything are still relatively prime, and (c) guarantees that it is composite.

After the iteration we should get $(a_1, \cdots, a_n)$ satisfy the condition we requested.

Answer 2

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 $x^6+y^6=2.$