Second degree polynomial solution existance Prove that if $p>2$ is prime and $p\equiv 2,3 \pmod5$, then congruence
$$x^2\equiv 5 \pmod{p}$$
doesn't have a solution.
Any ideas on how to approach this problem?
 A: Hint: Use Gauss's quadratic reciprocity.
$$
\left(\frac5p\right)=\left(\frac p5\right)
$$
Now observe that if $p\equiv r\pmod5$, then $\left(\frac p5\right)=\left(\frac r5\right)$. Now compute $\left(\frac r5\right)$ for a complete set of residues modulo $5$.
A: You may use the Legendre symbol if you're familiar with it,
and the Quadratic Reciprocity Theorem of Gauss.  
http://mathworld.wolfram.com/LegendreSymbol.html 
http://en.wikipedia.org/wiki/Legendre_symbol
Search here for: "For an odd prime p ≠ 5".    
http://mathworld.wolfram.com/QuadraticReciprocityTheorem.html 
Simply put, for any two distinct odd primes p and q, by using these techniques, one can transform the question of whether $x^2\equiv q \pmod{p}$ is solvable or not, to the question of whether $x^2\equiv p \pmod{q}$ is solvable or not. Note that by transform, I don't mean the answers of the two questions are always equal. The Quadratic Reciprocity Theorem from Gauss gives the link between the two answers. Again: here p and q are any two distinct odd primes.  
