Find a polynomial f(x) with real coefficients and f(0) = 1, such that the sums of the squares of the coefficients of f(x)n and (3x2 + 7x + 2)n are the same for all positive integers n.
Answer: 6x2 + 5x + 1.
The sums of the squares of the coefficients of anxn + an-1xn-1 + ... + a0 equals the coefficient of xn in (anxn + an-1xn-1 + ... + a0)(a0xn + a1xn-1 + ... + an). But (3x2 + 7x + 2)(2x2 + 7x + 3) = (6x2 + 5x + 1)(x2 + 5x + 6) = 6x4 + 35x3 + 62x2 + 25x + 6. So 6x2 + 5x + 1 has the required property.
How did we find it? Well, Note that 3x2 + 7x + 2 = (3x + 1)(x + 2), so we pick (3x + 1)(2x + 1).
46th Putnam 1985
© John Scholes
7 Jan 2001