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Discriminants (Posted on 2005-05-24) Difficulty: 2 of 5
Consider a quadratic equation with integer coefficients.
Is every integer a possible discriminant?

Prove it.

See The Solution Submitted by Jer    
Rating: 3.3333 (3 votes)

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Solution Puzzle Solution Comment 7 of 7 |

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
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