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Prove that there exist infinitely many integers $x$ such that integer $P(x)=x^3-2$ is a sum of two squares of integers.

Ideally, I am looking for a proof method that also applies for other $P(x)$, such as, for example, $P(x)=x^3+x+1$.

For $x=4t+3$, $P=(4t+3)^3-2$ is $1$ modulo $4$. By a well-believed (but difficult) Bunyakovsky conjecture, $P$ is a prime infinitely often, and every prime that is $1$ modulo $4$ is a sum of two squares.

To find an unconditional proof, it suffices to find polynomials $A(t)$, $B(t)$ and $C(t)$ with rational coefficients such that $A(t)^2 + B(t)^2 = C(t)^3-2$. Are there any good heuristics to find such polynomials? Is there any computer algebra system that helps to guess a solution of a polynomial equation over $Q[t]$?

One way to guess $C(t)$ is to form a set $S$ of all integers up to (say) $10^5$ that are sums of two squares, and look for polynomials $C(t)$ such that $C(t)^3-2$ belong to $S$ for all small $t$. I then checked all polynomials of degree up to $4$ and coefficients up to $12$, and found, for example, a polynomial $D(t)=3 + 8 t + 12 t^2 + 8 t^3 + 4 t^4$ such that $D(t)^3-1$ is always a sum of two squares. But no $C(t)$ found in this range, and increasing the degree and/or coefficients makes the enumeration infeasible. Are there methods to find a polynomial with all values in the given set, that are better than enumeration?

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  • $\begingroup$ As a side note, your $D(t)^3-1$ factors as $a(t) b(t) \overline{a}(t) \overline{b}(t)$ where $a(t) = t^2+2t+1+i$, $b(t) = 4t^4+8t^3+(12+2i)t^2+(8+2i)t+(3+2i)$, and the bar is complex conjugate. If you write $a(t) b(t) = u(t) + i v(t)$, then $u(t)^2+v(t)^2 = D(t)^3-1$. Unfortunately, I have no ideas about the main question. $\endgroup$
    – David E Speyer
    Commented Dec 2, 2021 at 17:39
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    $\begingroup$ One should be able to obtain a lower bound which tends to infinity, but not an asymptotic formula, for the number of integers $x$ such that $x^3 - 2$ is a product of at most $r$ primes (for some reasonable $r$, like $r = 11$ I think suffices), each of which is congruent to 1 mod 4. Each such number is then a sum of two squares. $\endgroup$
    – Stanley Yao Xiao
    Commented Dec 2, 2021 at 19:05
  • $\begingroup$ @ Stanley Yao Xiao - do you have any reference to this result? I found paper by H.-E. RICHERT "SELBERG'S SIEVE WITH WEIGHTS" that, under some natural conditions on P, proves that P(n) has a bounded number of prime factors infinitely often. However, how to add the condition that all these factors are congruent to 1 modulo 4? $\endgroup$
    – Bogdan Grechuk
    Commented Dec 13, 2022 at 18:40

2 Answers 2

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The answer is similar to one provided here. The approach is elementary and proves stronger fact that it can be expressed as sum of two coprime squares.

We consider the product $(n+2)(n^3-2)$ which is equal to $n^{4}-2n+2n^3-4$. Now we observe that $$(n+2)(n^3-2)=(n^2+n-7)^2+(13n^2+12n-53).$$ Proceeding as shown in the link we can easily get that $13n^2+12n-53$ is a perfect square for infinitely many $n\equiv 3\pmod{4}$ with first solution as $13(3^2)+12(3)-53=10^2$. If one chooses such a $n$ then clearly $n^3-2$ doesn't have any prime divisor of form $4k+3$ as $\gcd(n^2+n-7,13n^2+12n-53)\mid 25\cdot 59$ and $59$ doesn't divide $n^3-2$ if one looks at the general solution of the equation. Since, $n^3-2$ doesn't have any prime divisor of form $4k+3$ and $n\equiv 3\pmod{4}$ we can infer from the folklore result that it can be expressed as sum of two coprime squares.

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    $\begingroup$ Alternative way to prove along the same lines is to set $n=m^2+2$ and notice that $$(m^2+2)^3-2 = (m^3 + 3m)^2 + (3m^2 + 6).$$ Then solutions to $3m^2 + 6 = k^2$ (starting with $(m,k)=(1,3)$) will give required $n$. $\endgroup$
    – Max Alekseyev
    Commented Dec 2, 2021 at 19:26
  • $\begingroup$ Indeed @Max Alekseyev though I won't edit it now in hopes that me or someone is able to carry out the argument. $\endgroup$
    – Hhhhhhhhhhh
    Commented Dec 2, 2021 at 19:41
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    $\begingroup$ @MaxAlekseyev, yes, and then applying $m'=2m+k$, $k'=3m+2k$ gives larger solutions. $\endgroup$
    – user44143
    Commented Dec 2, 2021 at 20:35
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    $\begingroup$ Thank you, Max Alekseyev! This solution is much shorter than in the answer, and this idea should work for many other polynomials! $\endgroup$
    – Bogdan Grechuk
    Commented Dec 2, 2021 at 21:41
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One more way to solve the problem. Let $x = 4t + 3$. Then $$x^3 - 2 = 16t^2(4t + 9) + (108t + 25).$$ The system $$4t + 9 = a^2 \qquad 108t + 25 = b^2$$ has infinitely many solutions. It is reduced to the equation $b^2 - 27a^2 = - 218$. Also we can consider a system $$4t + 9 = ka^2 \qquad 108t + 25 = kb^2$$ with $k$ representable as a sum of two squares. It is reduced to the equation $b^2 - 27a^2 = - \frac{218}{k}$. There are solutions for $k = 109$.

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