I am currently attempting to prove that $b^n – 1$ is not a perfect square for any integers $b$ and $n$ greater than 1. I've gotten most of the way using a straightforward argument using quadratic residues modulo 4 (which are 0 for even squares and 1 for odd squares):
- If $b \equiv 0 \pmod{4}$, then $b^n – 1 \equiv 3 \pmod{4}$, which is not a quadratic residue.
- If $b \equiv 2 \pmod{4}$, then $b^n \equiv 2^n \equiv 0 \pmod{4}$ since $n \geq 2$, so $b^n – 1 \equiv 3 \pmod{4}$ again.
- If $b \equiv 3 \pmod{4}$, then $b^n \equiv (-1)^n \pmod{4}$. If $n$ is odd, then $b^n – 1 \equiv -2 \equiv 2 \pmod{4}$, which is not a quadratic residue. If $n$ is even, then $b^n$ is a perfect square, so $b^n – 1$ cannot be.
However, the case of $b\equiv 1 \pmod{4}$ still remains. Clearly, $b$ is odd. Further, if $b$ is square or $n$ is even, then $b^n$ is square and hence $b^n – 1$ is not square. I've checked the residues of $5^n – 1$ with moduli up to 1000 and none of them directly rule it out as a square, so I suspect I'm missing something (number theory is not my field). Does anyone have any suggestions?
Best Answer
I don't know of any way to solve your problem using congruence conditions, in particular dealing with your remaining case of $b \equiv 1 \pmod{4}$. Instead, consider if $b^n - 1$ is a perfect square, i.e., for some integer $m$ we have
$$b^n - 1 = m^2 \; \; \to \; \; b^n - m^2 = 1 \tag{1}\label{eq1A}$$
This is a special case of Catalan's conjecture, i.e., that
This was proven in $2002$ by Preda Mihăilescu. Since we have $m^2$ in \eqref{eq1A}, i.e., the exponent is $b = 2$ in the Catalan's conjecture equation, this shows there are no solutions to \eqref{eq1A}.
Note that \eqref{eq1A} was specifically solved earlier, as stated in Dietrich Burde's comment: