Prove that numbers divisible by $p$ but not by $p^2$ are quadratic non-residues of $p^n$

Question

Prove that numbers divisible by $p$ but not by $p^2$ are quadratic non-residues of $p^n$

This showed up in Disquisitiones Arithmeticae, article 102, but I fail to see why it must be true.

My attempt so far:

given $a\equiv 0\pmod p$, and $a\not\equiv 0\pmod {p^2}$ then $a = mp$, where m does not contain $p$ in its prime factorization. Then assume $a$ is a quadratic residue and look for a contradiction. $a\equiv b^2\pmod {p^n}$

$\frac {a-b^2}{p^n} = I$

$\frac {mp-b^2}{p^n} = I$

How should I proceed from there? Or is there another way to approach this?

Answer 1

Hint Show that $p|b^2$. Since $p$ is prime and $p|b^2$ you get that...

Answer 2

$$\implies a\equiv0\pmod p$$

If $x^2\equiv a\pmod p,x^2\equiv0\implies p |x^2\implies p|x\implies p^2|x^2\iff p^2|a$ which contradicts the given proposition

Now if there exists $y,y^2\equiv a\pmod{p^n}$ for integer $n>1$

$y^2\equiv a\pmod p$