For what integers n is the polynomial x4 - (2n + 4) x2 + (n - 2)2 the product of two non-trivial polynomials with integer coefficients?
Solution
Answer: n = square or twice a square.
We can write the polynomial as (x2 - (n - 2))2 - 8x2. This cannot have any integral solutions because then we would have A2 = 8 B2, and the lhs has an even number of powers of 2 and the rhs an odd number.
So if the polynomial factorises, it must do so as (x2 + ax + b)(x2 + Ax + B). Comparing terms in x3, A = -a. There are now two cases: (1) a is non-zero, (2) a is zero.
In case (1), comparing terms in x, we deduce B = b. Comparing constant terms, gives b = n - 2 or 2 - n. Comparing terms in x2 gives a2 = 2n + 4 + 2b. We cannot have b = 2 - n, for then a2 = 8, contradicting a integral. Hence, b = n - 2 and a2 = 4n. So n must be a perfect square. Put n = k2, then we can easily check that the polynomial factorises: x4 - (2k2+4)x2 + (k2 - 2)2 = (x2 +2kx + k2 - 2)(x2 -2kx + k2 - 2).
In case (2), we have b + B = -(2n + 4), bB = (n - 2)2. Hence b(-b - 2n - 4) = (n - 2)2, so b2 + b(2n + 4) + (n - 2)2 = 0. Hence (b + n + 2)2 = 8n. So n must be 2k2 for some integer k. It is easy to check that in that case the polynomial factorises: x4 - (4k2+4)x2 + (2k2 - 2)2 = (x2 - 2k2 - 4k - 2)(x2 - 2k2 + 4k - 2).
© John Scholes
jscholes@kalva.demon.co.uk
16 Dec 2001