Let $a$ be a quadratic residue modulo $p$. Prove that the number $bequiv a^frac{p+1}{4} mod p$ has the property that $b^2equiv a mod p$.

Let $p$ be a prime satisfying $p\equiv 3 \mod 4$. Let $a$ be a quadratic residue modulo $p$. Prove that the number $$b\equiv a^\frac{p+1}{4} \mod p$$ has the property that $b^2\equiv a \mod p$. (Hint: Write $\frac{p+1}{2}$ as $1+\frac{p-1}{2}$.) This gives an easy way to take square roots modulo $p$ for primes that are congruent to $3$ modulo $p$.
\textit{Hint:} Write $\frac{p+1}{2}$ as $1+\frac{p-1}{2}$ and use Exercise $3.36$.) This gives an easy way to take square roots modulo $p$ for primes that are congruent to $3$ modulo $p$.

I assume that the proof comes directly from the proof of quadratic residues but I am not sure how.

Answer

Yes, you are right, from the definition of $aR_p\implies a^{(p-1)/2}\equiv1\pmod p$

Now $\left(a^{(p+1)/4}\right)^2=a^{(p-1)/2}\cdot a\equiv a\pmod p$