Timeline for Is it true that $p^2+1$ is square free if $p>7$ is a Mersenne prime

7 events

when toggle format	what		by	license	comment
May 2, 2013 at 17:44	comment	added	Greg Martin		Indeed, given any $q\equiv1$ (mod $4$), the condition that $q^2 \mid (p^2+1)$ means $p^2\equiv-1$ (mod $q^2$), which is equivalent to $p$ lying in one of two residue classes modulo $q^2$. For example, when $q=5$, those residue classes are $7$ and $18$ (mod $25$). Solving $2^a-1\equiv7$ or $18$ (mod $25$) gives $a\equiv3$ or $18$ (mod $20$). Therefore $(2^a-1)^2+1$ is divisible by $5^2$ if and only if $a\equiv3$ or $18$ (mod $20$). While Mersenne primes are rare, nothing seems to keep them out of the residue class $3$ (mod $20$). Similar calculations hold for any $q\equiv1$ (mod $4$).
May 2, 2013 at 16:17	history	edited	Stefan Kohl♦	CC BY-SA 3.0	Added an answer to D. Burde's question.
May 2, 2013 at 15:20	comment	added	Barry Cipra		Just to flesh out Stefan's answer a bit, 25 is a divisor any time the exponent is congruent to 3 mod 20 (2 is a primitive root of unity mod 25), which is the case for 2203 and several other Mersenne primes.
May 2, 2013 at 11:37	comment	added	Dietrich Burde		Is $5$ here the only prime divisor occuring with multiplicity more than $1$ ?
May 2, 2013 at 10:09	vote	accept	BHZ
May 2, 2013 at 9:27	vote	accept	BHZ
May 2, 2013 at 10:09
May 2, 2013 at 9:23	history	answered	Stefan Kohl♦	CC BY-SA 3.0