Number of Solutions

[kathrynag]

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

 

[tkhunny]

It really is bad form to post in multiple locations:

http://www.mathhelpforum.com/math-help/f7/number-solutions-178816.html

Whose time is more valuable? Yours or volunteers who might help you? It's an interesting social dilemma.