Rohan
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 Dec5 awarded Popular Question Feb2 accepted show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ Feb2 comment show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ I get this but $a^2 \equiv k \pmod n$ and not necessarily $a^2 \equiv 1 \pmod n$ . Feb2 comment show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ Yes. the equation has a solution iff $p=4k+1$ Feb2 comment show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ ${{(-b^2)}^{2k+1}} \equiv 1 \pmod n \Rightarrow {b}^{2^{2k+1}} \equiv {-1} \pmod n$ which implies $1 \equiv {-1} \pmod n$ hence contradiction . It sounds fine to me. Is it good? Feb2 comment show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ This makes sense but we're not supposed to use a quadratic residue result yet. Maybe I should read the proof of this statement and try deriving it. Feb2 asked show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$ Jan31 comment Finding primes for which a given number is a perfect square. Yes, I was looking for a direction and not a solution . I enjoyed the severe reduction and limitations of the possibilities :) Jan31 awarded Supporter Jan31 awarded Scholar Jan31 comment Finding primes for which a given number is a perfect square. Thanks a lot. I got p=3,7 as the (only) two solutions Jan31 accepted Finding primes for which a given number is a perfect square. Jan31 awarded Student Jan31 awarded Editor Jan31 comment Finding primes for which a given number is a perfect square. sorry!, edited :) Jan31 revised Finding primes for which a given number is a perfect square. added 161 characters in body; edited tags Jan31 asked Finding primes for which a given number is a perfect square.