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The following problem is a $2000$ Hungarian Olympiad question. Find all primes $p$ such that:

$$p^n = x^3 + y^3$$

The answer is that there are only $2$ solutions:

  • $2^1 = 1^3 + 1^3$
  • $3^2 = 2^3 + 1^3$

Here's the argument:

  1. Assume $p \ge 5$ with $x,y,p,n$ positive integers and $p^n = x^3 + y^3$ where $n$ is the least possible value.
  2. Either $x$ or $y$ is greater than $1$ so that $x+y \ge 3$ and $xy \ge 2$
  3. $p^n = x^3 + y^3 = (x+y)(x^2 -xy + y^2)$
  4. $(x^2 -xy +y^2) = (x-y)^2 + xy \ge 2$
  5. Both $x+y$ and $x^2 - xy+y^2$ are divisible by $p$
  6. So, $p | (x+y)^2 - (x^2-xy+y^2) = 3xy$ and since $p \ge 5$, $p$ divides $x$ or $y$
  7. Since $p | x+y$, it follows that $p$ must divide both $x$ and $y$.
  8. So, $x^3 + y^3 \ge 2p^3$ and therefore $n > 3$
  9. But then we have a contradiction with step #1 since:

$$p^{n-3} = \left(\frac{x}{p}\right)^3 + \left(\frac{y}{p}\right)^3$$

Here's the possible generalization:

  1. Assume $p \ge 5$ with $x,y,p,n,m$ positive integers, $m$ odd, $m \ge 3$ and $p^n = x^m + y^m$ where $n$ is the least possible value.
  2. Either $x$ or $y$ is greater than $1$ so that $x+y \ge 3$ and $xy \ge 2$
  3. $p^n = x^m + y^m = (x+y)(x^{m-1} -x^{m-2}y + x^{m-3}y^{2} + \dots + -xy^{m-2} + y^{m-1})$
  4. $x^m + y^m - x - y = x(x^{m-1} - 1) + y(y^{m-1}-1) \ge 6$ so $\dfrac{x^m + y^m}{x+y} > 1$
  5. Both $x+y$ and $(x^{m-1} -x^{m-2}y - xy^{m-2} + \dots + (-xy)^{\frac{m-1}{2}} + y^{m-1})$ are divisible by $p$
  6. Since $p | (x+y)$, $x \equiv -y \pmod p$
  7. But this means that $p | y^{m-1}$ since $m$ is odd and there are $\frac{m+1}{2}$ powers of $x$ that are even and $\frac{m-1}{2}$ powers of $x$ that are odd with negative sign:

$$x^{m-1} -x^{m-2}y - xy^{m-2} + \dots + (-xy)^{\frac{m-1}{2}} + y^{m-1}$$

So, that we get:

$$y^{m-1} + y^{m-1} + y^{m-1} + y^{m-1} + \dots + y^{m-1} \equiv my^{m-1} \pmod p$$

  1. So, $p | y^{m-1}$ or $p | m$. If $p | y^{m-1}$, then $p | y$ and since $p | (x+y)$, $p | x$
  2. So, $x^m + y^m \ge 2p^m$ and therefore $n > m$
  3. But then we have a contradiction with step #1 since:

$$p^{n-m} = \left(\frac{x}{p}\right)^m + \left(\frac{y}{p}\right)^m$$

  1. For $p | m$, see here.

Thanks,

-Larry

Edit: A mistake was found in my argument. I'm updating. I'll try to fix it. If anyone has suggestions for whether the argument can be saved or not, I would appreciate it.

Edit 2: I believe that the argument for the generalization is now completed.

Edit 3: Made step 3 more standard based on a comment.

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Larry Freeman
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