I want to know whether my proof of $\mathbb{Q}_{p}\not\simeq \mathbb{Q}_{q}$ for $p\neq q$ primes is right or not. (I know that this question is already asked several times on MSE)
We can assume $q\neq 2$. We will try to find an integer $m$ such that $p\nmid m$ and $\left(\frac{mp}{q}\right) = 1$. (Legendre symbol). Then $mp$ is a square in $\mathbb{Q}_{q}$, but it is not in $\mathbb{Q}_{p}$ since the valuation is 1, which is odd.
If $\left(\frac{p}{q}\right) = 1$, then we can just set $m = 1$. If not, first choose $m'\in \mathbb{Z}$ with $\left(\frac{m'}{q}\right) = -1$. (Such $m$ exists since $q>2$). Then at least one of $m'$ and $m' + q$ is not a multiple of $p$, and choose $m$ as that one. Then $$ \left( \frac{mp}{q} \right) = \left( \frac{m}{q}\right)\left(\frac{p}{q}\right) = \left( \frac{m'}{q}\right)\left(\frac{p}{q}\right) = (-1)(-1) = 1, $$ so we are done. I think this proof works when $p =2$.