A question about Quadratic residues and modulos

Question

I need help in understanding why :

If a is a Quadratic residue modulo $p^l$ ( $\exists x$ such that $x^2 \equiv a \mod(p^l)$ ) then its a Quadratic residue modulo p ($ \exists x$ such that $x^2 \equiv a \ mod (p)$ ).

Answer 1

One usual definition of quadratic residue goes as follows: $a$ is a quadratic residue of $m$ if $\gcd(a,m)=1$ and there exists an $x$ such that $x^2\equiv a\pmod{m}$.

Suppose that $a$ is a quadratic residue of $p^l$, where $l\ge 1$. Then $\gcd(a,p^l)=1$ and there is a $b$ such that $b^2\equiv a\pmod{p^l}$.

It follows that $\gcd(a,p)=1$. Furthermore, since $p^l$ divides $b^2-a$, it follows that $p$ divides $b^2-a,$ and therefore $b^2\equiv a\pmod{p}$. Thus by definition $a$ is a quadratic residue of $p$.

Note that, in the above argument, there is no need to assume that $p$ is prime.

Remark: It is more challenging to work in the opposite direction, and try to show that if $a$ is a QR of $p$, then $a$ is a QR of $p^l$. This is true for all odd primes $p$.

Answer 2

Put $N=Mp^{l-1}$ $$x^2=a+Mp^l\Rightarrow x^2=a+Np$$