0

From our lecture, I noted down this statement: Let $a$ be an integer and $m$ an odd natural number with the prime decomposition $$ m = p_1^{\alpha_1}\cdot p_2^{\alpha_2} \cdots p_n^{\alpha_n}. $$ If $a$ is a quadratic residue mod $m$, then $a$ is quadratic residue for every divisor of $m$.

We also have a proof and it is an easy one: Let $a$ be a quadratic residue mod $m$. Then we have an integer $x$ sch that $$ a\equiv x^2 \pmod m \Leftrightarrow m\mid a-x^2 \Rightarrow p\mid a-x^2 \Leftrightarrow a\equiv x^2 \pmod p $$ for every divisor $p$ of $m$.

But I am wondering why we did this for an odd $m$. It should also be possible for even numbers $m$, and the proof that we have does not distinguish. What am I missing?

Lereu
  • 424
  • 1
    This specific statement is true for even $m$ as well, with the same proof. But most of the theory about quadratic residues fails for the even numbers, which I guess is why they are not considered here. – Mark Nov 14 '23 at 10:33
  • @Mark Thank you. I added the proof now in my post. – Lereu Nov 14 '23 at 10:43
  • 2
    Actually, your proof is somewhat invalid. You should say that $$\forall i \in {1,2,\cdots,n},~ m ~\mid ~a-x^2 ~\implies ~(p_i)^{\alpha_i} ~\mid ~a-x^2.$$The point is that $~a~$ is a quadratic residue for any divisor of $~m.$ – user2661923 Nov 14 '23 at 10:50
  • @user2661923 Hm I guess I should have noted down more things. The $p_1$ to $p_n$ don't have be different in my statement. – Lereu Nov 14 '23 at 11:23

0 Answers0