16th Irish 2003 Problem B3 Find all integer solutions to m 2 + 2m = n 4 + 20n 3 + 104n 2 + 40n + 2003.
m = 128, n = 7 or -17. Solution Note that m 2 + 2m + 1 = (m + 1) 2 and n 4 + 20n 3 + 104n 2 + 40n + 4 = (n 2 + 10n + 2) 2 . Put M = m+1, N = n 2 + 10n + 2 and the equation becomes M 2 - N 2 = 2000. Thus (M + N)(M - N) = 2000, also M + N and M - N have the same parity (since their sum is even). They cannot both be odd, since 2000 is even, so they must both be even. Thus {M + N, M - N} must be {2, 1000}, {4, 500}, {8, 250}, {10, 200}, {20, 100} or {40, 50} or the corresponding negative pairs. But note that N = (n + 5) 2 - 23 >= -23. Thus N = (1000 - 2)/2 = 499, 248, 121, 95, 40, 5 or -5. That corresponds to (n+5) 2 = 522, 271, 144, 118, 63, 28, 18. So the only possibilitiy is 144, giving n = 7 or -17 and N = 121, M = 129, m = 128. 16th Irish 2003 © John Scholes jscholes@kalva.demon.co.uk 14 June 2003 Last updated/corrected 14 Jun 03