Using the Legendre Symbol Which have solutions? $x^{2} \equiv7 \mod{53}$,  $x^{2} \equiv53 \mod{7}$,  $x^{2} \equiv 14 \mod{31}$, $x^{2} \equiv 25\mod{997}$?
I have all of these properties for this Legendre symbol and no idea what to do with it to find whether these have a soltion, let alone solve them if they do. 
 A: They all have solutions. Since $53 \equiv 1 \pmod{4}$, we have
$$\left(\frac{7}{53}\right) = \left(\frac{53}{7}\right) = \left(\frac{4}{7}\right) = \left(\frac{2}{7}\right)^2 = 1,$$
which settles the first two. $31 \equiv 7 \pmod{8}$, so
$$\left(\frac{14}{31}\right) = \left( \frac{2}{31}\right) \left(\frac{7}{31}\right) = \left(\frac{7}{31}\right) = - \left(\frac{31}{7}\right) = - \left(\frac{3}{7}\right) = \left(\frac{7}{3}\right) = \left(\frac13\right) = 1.$$
And finally
$$\left(\frac{25}{997}\right) = \left(\frac{5}{997}\right)^2 = 1.$$
Finding solutions is a different matter, but
$$22^2 = 484 = 9\cdot 53 + 7;\quad 2^2 = 4 \equiv 53 \pmod{7};\quad 13^2 = 169 = 5\cdot 31 + 14; \quad 5^2 = 25.$$
A: For the first, we want $(7/53)$. Both of our numbers are primes, and one has form  $4k+1$. So by Quadratic Reciprocity, $(7/53)=(53/7)$.  
But $53\equiv 4\pmod 7$, so $(53/7)=(4/7)$. But $4$ is a perfect square, so $(4/7)=1$. It follows that $(7/53)=1$, so $x^2\equiv 7\pmod{53}$ has a solution. 
While solving the first problem, we solved the second. We saw that $(53/7)=1$, so the congruence $x^2\equiv 53\pmod{7}$ has a solution. But we really need minimal machinery for that, since $53\equiv 4\pmod{7}$, so the congruence has the obvious solutions $x\equiv \pm 2\pmod{7}$.
Jumping ahead, the congruence $x^2\equiv 25\pmod{997}$ has at least one obvious solution, no Legendre symbol calculation needed.
For the congruence $x^2 
\equiv 14\pmod{31}$, the Legendre symbol approach has us evaluate $(14/31)$.  By one of the standard properties of Legendre symbols, this is $(2/31)(7/31)$.
Because $31$ is a prime of the shape $8k+1$, we have $(2/31)=1$. Now we need to evaluate $(7/31)$. Since each of $7$ and $31$ is a prime of the form $4k+3$, by Reciprocity argument we $(7/31)=-(31/7)$. But $31\equiv 3\pmod{7}$, so we want $-(3/7)$. By Reciprocity, this is the negative of the negative of $(7/3)$, so it is $(7/3)$. But $(7/3)=(1/3)=1$. It follows that the congruence $x^2\equiv 14\pmod{31}$ has a solution.
