Using the Law Quadratic Residues

Question

I'm looking to find if $\bigg(\frac{1210}{41}\bigg)$ so is 1210 a quadratic residue modulo 41.

So I have $\bigg(\frac{1210}{41}\bigg)=\bigg(\frac{21}{41}\bigg)=\bigg(\frac{3}{41}\bigg)\bigg(\frac{7}{41}\bigg)$

Taking the question in parts i get $\bigg(\frac{3}{41}\bigg)\bigg(\frac{41}{3}\bigg)=(-1)^{20}=1 \rightarrow \bigg(\frac{2}{3}\bigg)=-1$

Using this i then have $-\bigg(\frac{7}{41}\bigg)=-\bigg(\frac{7}{41}\bigg)\bigg(\frac{41}{7}\bigg)=(-1)^{60}=1 \rightarrow -\bigg(\frac{41}{7}\bigg)=-\bigg(\frac{6}{7}\bigg)=1$

So i get -1 therefore 1210 is not a Quadratic Residue modulo 41. Is this the correct answer for this question and correct method?

Answer 1

$$\left(\frac3{41}\right)=\left(\frac{41}3\right)=\left(\frac23\right)=-1$$

$$\left(\frac7{41}\right)=\left(\frac{41}7\right)=\left(\frac{-1}7\right)=-1$$