Oops, I typed out the solution to problem 1 instead of problem 2. For problem 2, you do exactly the same thing. Since the  modulii are not  prime, quadratic residues must be relatively prime to the modulus.

1. There are several ways to determine the quadratic residues. Either you can square the least positive residues and see what results - or - you can compute the Legendre symbol (a/p) and the quadratic residues will correspond to 1. I'll take the first approach (mostly because it is easier to type!)

2(c). You will find that only 1 and 4 are quadratic residues modulo 15.
2(d). You will find that only 1, 7, and 13 are quadratic residues modulo 18.

5.(a) To evaluate the Legendre symbol (7/11) using Euler's criterion we compute

   (b). Using Gauss' Lemma, we first need to find how many of the least positive residues listed below are greater than 11/2=5.5.
 

There are 3 least positive residues greater than 5.5, so the Legrendre symbol (7/11) = (-1)³ = -1.

7. The Legendre symbol (-2/p) = (-1/p)(2/p).

If p1 mod 8, then (-2/p) = (1)(1) = 1.
If p3 mod 8, then (-2/p) = (-1)(-1) = 1.
If p-1 mod 8, then (-2/p) = (-1)(1) = -1.
If p-3 mod 8, then (-2/p) = (1)(-1) = -1.

So (-2/p)= 1 whenever p1 mod 8 or p3 mod 8 and (-2/p)= -1 for  p-1 mod 8 or p-3 mod

8.

13. Let's use the discriminant (as described in problem 12) to determine how many solutions to expect for each problem. Then it's time to dust off our factoring caps.

(a). d = b²-4ac = 1 -4(1)(1)= -34 mod 7. An 4 is definitely a quadratic residue modulo 7, so we expect 2 incongruent solutions. Notice that  x²+x+1  (x -2)(x +3) mod 7. So the solutions to x²+x+1  0 mod 7 are x  2 mod 7 and x  -3 mod 7 (or  x  4 mod 7).

(b). d = b²-4ac = 25 -4(1)(1)= 21 0 mod 7.
So we expect a single solution.  Notice that (x + 6)² x² + 5x + 1 mod  7. So the solution to x² + 5x + 10 mod 7  is x  6  mod 7.

19. Any solution to x² 1 mod 15 must simultaneously satisfy  x² 1 mod 3 and x² 1 mod 5.

The solutions modulo 3 are x(1 or 2) mod 3 and the solutions modulo 5 are x(1 or 4) mod 5.
This gives us 4 posssible solutions. Use the Chinese Remainder Theorem to find them all.
 

22 .  If x₀ is a solution to x² a (mod p^e), then (- x₀)²a (mod p^e) and  x₀ does not equal - x₀ since  p^e  does not divide 2x₀. Si if there is one solutions, there must be two. Now suppose that x₀ and x₁ are solutions. Then x₀²x₁²a (mod p^e). So  p^e |(x₀²-x₁²)=(x₀ -x₁)(x₀ + x₁).  Notice that p cannot divide both factors because then p| (x₀ -x₁) + (x₀ + x₁) = 2x₀ which is impossible since (a,p)=1. So we must have either p^e | (x₀ -x₁)   or  p^e | (x₀ + x₁). Hence x₀ +/- x₁ mod  p^e. So there are at most 2 solutions.