### 40th Putnam 1979

**Problem B3**

F is a finite field with n elements. n is odd. x^{2} + bx + c is an irreducible polynomial over F. For how many elements d ∈ F is x^{2} + bx + c + d irreducible?

**Solution**

Since n is odd, h ≠ -h for any h ∈ F. So there are exactly (n + 1)/2 quadratic residues. Since n is odd, we may write b as 2k, and so we are interested in d for which (x + k)^{2} = k^{2} - c - d is irreducible. In other words, d for which k^{2} - c - d is a quadratic non-residue. But k^{2} - c - d runs through all the elements of F, so there are just (n - 1)/2 values of d which give a quadratic non-residue.

40th Putnam 1979

© John Scholes

jscholes@kalva.demon.co.uk

4 Dec 1999