Diary

Diary 2026 08 30 Learning some new techs from Richard’s online courses. Tech 1 Given that, $$ p | (n^2+1) $$ We have, $$ n^2 \equiv -1 \pmod p \\ n^4 \equiv 1 \pmod p $$ So $4|p-1$ or $p=2$ (in which case $-1 = 1$) Tech2 Given that, $$ p | (a^q-1), p \not\mid (a-1) $$ where $p,q$ are primes. We learn that, $$ a^q \equiv 1\pmod p $$ So the order of $a$ could only be $q$ or $1$. We also know from $p \not\mid (a-1)$, that the order is not 1. ...