Consider a quadratic equation with integer coefficients.
Is every integer a possible discriminant?

Prove it.

Let the quadratic equation in x be given by p*x^2 + qx + r, where each of p, q and r is an integer.

Then the discriminant(D) = q^2 - 4pr
If  q^2 = 4pr, then D=0
If  q^2 != 4pr, then for odd q, we have q^2(mod 8) = 1, so that: q^2 has the form: 8g+1, for an integer g.
So, D = q^2 - 4pr = 8g + 1 - 4pr = 4(pr+2g) + 1, so that:
D (mod 4) = 1

If  q^2 != 4pr, then for even q,we have q^2(mod 4) = 0, so that: q^2 has the form: 4h, for an integer h.
So, D = q^2 - 4pr = 4h - 4pr = 4(pr+h), so that:
D (mod 4) = 0

Consequently, D = 0, and:

D(mod 4) = 0, or 1 whenever D is nonzero. 

Posted by K Sengupta on 2009-01-22 13:58:12