I came across a math olympiad type question that goes like this:

For what primes $p$ will $p=m^2+n^2$ and $p$ divide $m^3+n^3-4$?

I tried a few examples and think that $p=5$ is the only solution but am unable to prove it. The property that $p \equiv 1 \mod 4$ seems to not be helpful so far. Any clues or solutions are welcome.

  • $\begingroup$ You have $m^3+n^3-4$ in the title but exponents are $2$ in the body. Which is correct? (I guess the former) $\endgroup$ – Wojowu Sep 1 '16 at 18:50
  • $\begingroup$ Note: I reformatted the title but left the body the same (though I am fairly sure you meant to have $m^3+n^3-4$ in the body as well). If that's correct, please edit accordingly. $\endgroup$ – lulu Sep 1 '16 at 18:51
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    $\begingroup$ $p = 2$ technically works too. $\endgroup$ – 6005 Sep 1 '16 at 19:11
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    $\begingroup$ Are $m$ and $n$ positive? $\endgroup$ – 6005 Sep 1 '16 at 19:12
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    $\begingroup$ Just an interesting course of events: $$p=m^2+n^2$$ $$mp=m^3+mn^2$$ $$np=nm^2+n^3$$ $$p(n+m)-(nm^2+mn^2)-4=m^3+n^3-4$$ $$p(n+m)-nm(m+n)-4=m^3+n^3-4$$ $$nm(m+n) \equiv -4 \pmod{p}$$ $$n^2m^2(m^2+n^2+2mn) \equiv 16 \pmod{p}$$ $$2n^3m^3 \equiv 16 \pmod{p}$$ $p=2$-one solution, otherwise: $$n^3m^3 \equiv 8 \pmod{p}$$ $\endgroup$ – rtybase Sep 1 '16 at 21:45

Not an answer!!! Work in progress.

We can use the Gaussian integers. Assume that $m^2 + n^2$ divids $m^3 + n^3 - 4$.

Note that $m^2 + n^2 = (m + in)(m - in)$, and since $m^2 + n^2$ is prime in $\mathbb{Z}$, both of these are prime in $\mathbb{Z}[i]$. Additionally, by this result, as $m,n$ are relatively prime here, $\mathbb{Z}[i] / (m+in) \cong \mathbb{Z}/(m^2 + n^2) \mathbb{Z}$. So we work in the field $\mathbb{Z}[i] / (m+in)$, getting $$ m^3 + n^3 - 4 \equiv (-in)^3 + n^3 - 4 = (i+1)n^3 - 4 \equiv 0 \pmod{m+in} $$ Thus $$ (1+i) n^3 \equiv -4 = -(1+i)^4 \pmod{m+in}. $$ If $m = n = 1$, we have a solution $\boxed{p=2}$. Otherwise, $1+i$ is not zero, and we divide by it to get $$ n^3 = -(1+i)^3 \pmod{m+in}. $$

So the question now is to solve a cubic in $\mathbb{Z}[i]/(m+in)$. It definitely has one solution, $n \equiv -(1 + i)$. It may have two other soutions, if $1$ has three cube roots in $\mathbb{Z}[i](m+in)$, which happens exactly when $m^2 + n^2 \equiv 1 \pmod{3}$ (see here). If so, then there is a Gaussian integer $\omega$, $\omega \ne 1$, such that $\omega$ and $\omega^2$ are the two nontrivial cube roots of $1$ in $\mathbb{Z}[i] / (m+in)$. Otherwise, let $\omega = 1$. Either way, we have $n \equiv -\omega^a (1+i) \pmod{m+in}$, for $a \in \{0,1,2\}$.

Using the isomorphism between $\mathbb{Z}[i]/(m+in)$ and $\mathbb{Z}[i]/(m-in)$ given by conjugation, of course $n \equiv -(\overline{\omega})^a (1-i) \pmod{m-in}$.

  • $\begingroup$ In your first line, do you mean "divides $m^3+n^3 - 4$"? $\endgroup$ – Erick Wong Sep 1 '16 at 20:22
  • $\begingroup$ @ErickWong Yes, thanks. $\endgroup$ – 6005 Sep 1 '16 at 20:23

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