let $a,b$ be integers, such that for every positive integer $n$, the expression :

$[(a^n)-1][(b^n)-1]$ is a perfect square. How to prove that $ab$ is a perfect square.

  • 3
    $\begingroup$ What about $a=1$, $b=2$? $\endgroup$ – Erick Wong Jan 10 '13 at 8:44
  • $\begingroup$ Sorry ,$a$ and $b$ are greater than one. $\endgroup$ – user56777 Jan 10 '13 at 8:54

Since $[a^{3n}-1][b^{3n}-1]$ is a square and $[a^{n}-1][b^{n}-1]$ is a non-zero square, the quotient $[a^{2n}+a^n+1][b^{2n}+b^n+1]$ is also a square for all $n\in\mathbb N$.

Suppose that $ab$ is not a square. A consequence of quadratic reciprocity is that there are infinitely many primes $p$ such that $\left(\frac{ab}p\right) = \left(\frac{3}p\right)=-1$ (the density of such primes is at least $\tfrac14$). Fix $p$ to be one such prime.

Now, exactly one of $a$ and $b$ is a quadratic residue mod $p$. Choosing $n=(p-1)/2$, Euler's criterion shows that $a^{2n}+a^n+1$ is $3 \pmod p$ if $a$ is a residue and $1\pmod p$ if $a$ is a non-residue. Therefore $[a^{2n}+a^n+1][b^{2n}+b^n+1] \equiv 3 \pmod p$, so it can't be a square.

This is a very nice problem, and I imagine that other quite different attacks are possible. I'd like to know if there is a way to deduce the much stronger statement that $a = b$, since that seems like the only plausible way to make $[a^{n}-1][b^{n}-1]$ always a non-zero square.

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    $\begingroup$ how do you prove the existence of such prime $p$ ? $\endgroup$ – user56777 Jan 10 '13 at 9:46
  • $\begingroup$ @user56777 That's fairly standard, but takes a bit of fiddling. Reciprocity shows that if $q_1, q_2, \ldots, q_k$ are distinct primes, then you can independently prescribe each $(q_i/p)$ by choosing $p$ in the right congruence class. Now choose $(3/p) = -1$ and (if needed) some other prime $q$ dividing $ab$ to have $(q/p)=-1$. $\endgroup$ – Erick Wong Jan 10 '13 at 9:56
  • $\begingroup$ @ErickWong: How does that imply that $ab \equiv 3 \pmod p$? $\endgroup$ – Sandeep Silwal Jan 20 '15 at 15:18
  • $\begingroup$ @SandeepSilwal I never claimed $ab \equiv 3 \pmod p$, nor is it expected by the problem. It appears you are confused about the meaning of the Legendre symbol: en.wikipedia.org/wiki/Legendre_symbol $\endgroup$ – Erick Wong Jan 20 '15 at 15:32

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