Let $p\equiv3\pmod4$ be prime and $p \nmid x$.
How can you show that a quadratic residue modulo $p$ is given by $x$ or $-x$ ?
Thanks!
Let $p\equiv3\pmod4$ be prime and $p \nmid x$.
How can you show that a quadratic residue modulo $p$ is given by $x$ or $-x$ ?
Thanks!
Hint $ $ mod $\rm\, P = 4K\!+\!3\!:\ \color{#C00}{(-x)^{(P-1)/2}}\, \color{#0A0}{x^{(P-1)/2}} \equiv (-1)^{2K+1} x^{P-1} \equiv -1\ $ thus $\rm\color{#C00}{one}\ \color{#0A0}{factor}$ is $\equiv 1.\,$
Hence by Euler's Criterion, $\rm\, \color{#C00}{-x}\ $ or $\rm\ \color{#0A0} x\ $ is a square $\rm\, mod\ P.\,$
Remark $\ $ A similar argument using Euler's criterion allows one to prove a general formula for the product of (non)residues, encapsulated as Legendre symbol multiplicativity$\rm\,\ (xy|p) = (x|p)\,(y|p).$ The above is the special case $\rm\:y = -x,\:$ using $\rm\:(-1|p) = -1\:$ for primes $\rm\:p\equiv 3\pmod 4.$
Hint. If $p\equiv 3\pmod{4}$, then $-1$ is not a square modulo $p$. If you multiply two non-squares modulo $p$, what do you get?