Funny Diophantine from 2023 RMM, Problem 1

投稿日:2023/2/3最終更新日:2025/5/18

I hate combinatorics 😡

Problem Statement

Determine all prime numbers $p$ and all positive integers $x$ and $y$ satisfying

$$x^3 + y^3 = p(xy + p).$$

Solution

Surprising

The solutions are $(x,y,p)=(8,1,19),(7,2,13),(5,4,7)$, and the obvious $3$ others by swapping $x$ and $y$.

WLOG let $x \geq y$. Note that $p \nmid x,y$, since if it divides one then it divides the other, but then $\nu_p(LHS)\geq 3$ while $\nu_p(RHS)=2$. Furthermore, no prime $q \neq p$ divides both $x$ and $y$, else it divides $p$, which is absurd. Now we observe that if $x>p$, then we have

$$x^3+y^3 < x(xy+x) = x^2y + x^2 \implies x^2(x-y-1)+y^3 < 0,$$

which is also absurd, as this inequality is clearly false for $x=y=1$ and otherwise $x\geq y-1$ since $x \neq y$ is impossible. Thus $x \leq p$, so in fact $p>x>y$.

Factor the LHS as $(x+y)(x^2-xy+y^2)$. If $p \mid x+y$, then $x+y=p$ for size reasons. But then $x^2-xy+y^2=xy+p \implies (x-y)^2=p$, which clearly never happens. Thus $p \mid x^2-xy+y^2$, so write $x^2-xy+y^2=kp$. We then have

$$(x+y)k=xy+p \implies (x+y)k^2-kxy=(x+y)^2-3xy \implies (x+y)(k^2-x-y)=(k-3)xy.$$

Clearly $x+y \mid (k-3)xy$. Because $\gcd(x,x+y)=\gcd(y,x+y)=1$, it follows that $x+y \mid k-3$. Suppose that $k \neq 3$. Because $x$ and $y$ are positive we can't have $k=2$, and if $k=1$ then $x=y=1$ which is clearly never a solution for any primes $p$. Thus $k-3$ is positive and thus at least $x+y$. Substituting this back into the original definition of $k$, we have

$$x^2-xy+y^2\geq (x+y+3)p>x(x+y+3) \implies y^2>2xy+3x,$$

which is preposterous. Hence $k=3$, so we should also have $x+y=k^2=9$. A case check finishes.

Remark Although this solution does not appear to be heavily motivated by the equality cases, it was. One notices that the value of $k$ must always equal $3$, which makes it inevitable that the term will be used somehow. On the other hand, the mere existence of solutions implies that one cannot blindly go in applying size or some other method to obtain a contradiction. When it comes to finding solutions themselves, this is much easier once we reduce the problem to $p \mid x^2-xy+y^2$. For cyclotomic polynomial reasons this implies that only $p \equiv 1 \pmod{6}$ possibly work, and for those primes $p$ there is some integer $a$ such that any solution $(x,y)$ will have $x \equiv ay \pmod{p}$ up to switching $x$ and $y$ (namely $a$ such that $a^2-a+1 \equiv 0$), so our search space is greatly narrowed. I would be surprised to find a solution that relies on $p \equiv 1 \pmod{6}$ being true.