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It is known that if $x^2 + y^2 = z^2$ is a primitive Pythagorean triplet then $z$ is not divisible by any prime of the form $4k-1$. The following is a generalization of this classical result which shows that the source of this property is not the hypotenuse $z$, but the two orthogonal side $x$ and $y$:

Conjecture: Let $f(x,y) = a_0x^n + a_1 x^{n-1}y + a_2x^{n-2}y^2 + \cdots + a_ny^n$, $a_0a_n \ne 0$. Then there are infinitely many primes of the form $8k+3$ which do not divide $f(x,y)$ for any primitive Pythagorean triplet $x^2 + y^2 = z^2$.

Example: Taking $f(x,y) = x^n + y^n$, the conjecture says that for every there are infinitely many primes which do not divide the sum of the two orthogonal sides of any Pythagorean triangle. The conjecture has been proved for $x+y$ (in MSE) and is already known to be true for $x^2 + y^2$. Experimental data shows that $x^3 + y^3$ is not divisible by infinitely many primes of the form $8k+3$ while $x^4 + y^4$ is not divisible by infinitely many primes of the form $8k+3, 8k+5$ and $8k+7$.

Question: Is the conjecture known? I am looking for a proof or disproof of the conjecture.

The conjecture was proved for special case $f(x) = x+y$ in MSE but the general case remains open. Hence it is posted in MO.

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  • $\begingroup$ Some of the binomials were studied by Legendre who had divisibility results that depended on the exponent, likely motivated by the (then) Fermat conjecture. Gerhard "Which You May Already Know" Paseman, 2020.07.05. $\endgroup$
    – Gerhard Paseman
    Commented Jul 5, 2020 at 16:32
  • $\begingroup$ Also, what about x^2 + 3y^2? Gerhard "A Reason For Being Unknown?" Paseman, 2020.07.05. $\endgroup$
    – Gerhard Paseman
    Commented Jul 5, 2020 at 16:39
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    $\begingroup$ @GerhardPaseman Since $x^2 + 3y^2 = (x^2 + y^2) + 2y^2$ hence trivially the conjecture is true for $x^2 + 3y^2$ since it is true for $x^2 + y^2$ $\endgroup$
    – Nilotpal Kanti Sinha
    Commented Jul 5, 2020 at 18:22
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    $\begingroup$ Perhaps I misunderstand the question, but doesn't $f(x,y) := 4x-3y$ give a counterexample just because of the Pythagorean triple $(x,y,z)=(3,4,5)$? Or if one wants a nonnegative example, take $f(x,y) = 4x+3y$ and use the triples $(x,y,z) = ((p-2)^2-1, 2(p-2), (p-2)^2+1)$ (which is congruent mod $p$ to $(3,-4,5)$) for a given prime $p = 8k+3$. $\endgroup$
    – Terry Tao
    Commented Jul 6, 2020 at 3:38

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