Show that for any integers a, b one can find an integer c such that there are no integers m, n with m2 + am + b = 2n2 + 2n + c.
Solution
If a is odd, then we just take c to have the opposite parity to b, since m2 + am + b = m(m+a) + b has the same parity as b (for all m), and 2n2 + 2n + c has the same parity as c (for all n).
Suppose a is even. Then 2n2 + 2n + c = 2n(n+1) + c = c mod 4. But m2 + am + b = (m + a/2)2 - a2/4 + b. Any square is 0 or 1 mod 4, so m2 + am + b must be -a2/4 + b mod 4 or 1 - a2/4 + b mod 4. Thus we can take c to be any number which is not -a2/4 + b mod 4 and is not 1 - a2/4 + b mod 4.
© John Scholes
jscholes@kalva.demon.co.uk
8 Oct 2002