# Kronecker symbol and the connection to quadratic reciprocity

I am reading about kronecker symbol, and I have read that this simple has no connection to quadratic residues. This make me very confused since quadratic reciprocity very related to quadratic residues since it depends on the definition of Quadratic residues. Whereas the Kronecker symbol has a quadratic reciprocity law. But then what does the quadratic reciprocity for kronecker symbol say about the solvability of the congruence $$x^2 \equiv m \mod n$$ for positive integers $$m$$ and $$n$$ ( not necessarily odd when applying kronecker symbols)? and what is the importance of this symbol?

• Let $s(m \bmod n)=1$ if $\gcd(m,n)=1$ and $\exists x, m \equiv x^2 \bmod n$, $s(m \bmod n)=0$ otherwise. If I'm not wrong $s(m \bmod n) = \prod_{p^k | n} s(m \bmod p^k)$, for $p$ an odd prime $s(m \bmod p^k)= s(m \bmod p)$. – reuns Nov 28 '18 at 0:59
• Where did you read that "the Kronecker symbol has no connection to quadratic residues"? – Rob Arthan Nov 28 '18 at 1:32
• @Rob Arthan this is a statement from Wikipedia "On the other hand, the Kronecker symbol does not have the same connection to quadratic residues as the Jacobi symbol. In particular, the Kronecker symbol ${\displaystyle \left({\tfrac {a}{n}}\right)}$ for even n can take values independently on whether a is a quadratic residue or nonresidue modulo n" Rob can you explain if you know any connection to quadratic reciprocity? – Rosa Nov 28 '18 at 3:37
• Thank you @reuns . Again I can't see any connection. Maybe because I am not familiar with your notations. – Rosa Nov 28 '18 at 3:40
• Your question is about the value of $s(m \bmod n)$. Yes it isn't directly related to $(\frac{m}{n})$ for $n$ not a prime. – reuns Nov 28 '18 at 4:31