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?
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