For each of the following congruences, determine the number of solutions.

(a). x^2 =103 (mod 420).

(b). x^2 =29 (mod 1080).

(c). 2x^2 -5x + 23 = 0 (mod 675).

(d). x^2 + 3x -22 = 0 (mod 108).

Ideas:

a)x^2=103 mod(420)

I notice that 420=2^2*3*5*7

So we are interested in N(240)=N(2^2)*n(3)*N(5)*N(7)

Now N(2^2)=0 103=3(mod 4)

so 0 solutions

b)x^2=29 (mod 1080)

I notice that 1080=2^3*3^3*5

So we are interested in N(1080)=N(2^3)*N(3^3)*N(5)

That's as far as I get

c)2x^2 -5x + 23 = 0 (mod 675).

We can begin by finding the discriminant

(-5)^2-4(2)(23)

25-184=-159

So we will solve x^2=-159 (mod 675)

Is it not solvable since -159 is not relatively prime to 675?

d)x^2 + 3x -22 = 0 (mod 108).

We can find the discriminant

3^2-4(1)(-22)

9+88=97

We will solve x^2=97 (mod 108)

Notice 108=2^2*3^3

Consider N(108)=N(2^2)*N(3^3)

That's as far as I get