show if n=4k+3 is a prime and a2+b2≡0(modn) , then This question has already been posted here.
show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$
And it also has satisfactory answers however I have not yet studied quadratic residues and though the other answers are well written and I have understood them , my book has given a hint which goes this way : 
"If a is not congruent to 0 modulo p then there exists an integer 'c' such that $$ ac\equiv 1\pmod n $$ 
Any help in approaching this problem using the above hint is appreciated
PS: Please do not mark this as duplicate as I want to understand the usage of this hint in solving the above problem

You already got one good answer. The following alternative proof is more elementary.
Assume the claim does not hold. Let $n,a,b$ be a counterexample such that $n$ is as small as possible. By the hint,
$$(bc)^2+1\equiv(bc)^2+(ac)^2\equiv(a^2+b^2)c^2\equiv 0\pmod{n}.
$$
Let $d$ be an even number such that $0<d<n$ and $d\equiv\pm bc\pmod{n}$. Then $d^2+1=mn$ for some positive integer $m<n$. Now $m\equiv 3\pmod{4}$, and therefore $m$ has a prime factor $p$ such that $p\equiv 3\pmod{4}$. Then $d^2+1^2\equiv 0\pmod p$, contradicting the minimality of $n$.
