Show that we can find integers a, b, c, d such that a2 + b2 + c2 + d2 = abc + abd + acd + bcd, and the smallest of a, b, c, d is arbitarily large.
If b, c, d are fixed, then the relation gives a quadratic in a whose two roots have sum bc + bd + cd. So if we have a solution a, b, c, d, then we can derive another solution a', b, c, d with a' = (bc + bc + cd) - a (*). If we take a < b < c < d, and all of a, b, c, d positive, then clearly a' > d.
So starting with any positive solution we may derive solutions with successively larger smallest members. In fact, starting with (1, 1, 1, 1), we get successively (1, 1, 1, 2), (1, 1, 2, 4), (1, 2, 4, 13), (2, 4, 13, 85). From this point on, the smallest member is increased each time.
39th Putnam 1978
© John Scholes
30 Nov 1999