Show that x4 - 1993 x3 + (1993 + n) x2 - 11x + n = 0 has at most one integer root if n is an integer.
Solution
Solution by John R Ramsden
If it has two integer roots, then a quadratic (x2 - ax + b), with a and b integral, is a factor of x4 - 1993 x3 + (1993 + n) x2 - 11x + n. Suppose the other factor is (x2 - cx + d). So (x2 - ax + b)(x2 - cx + d) = x4 - 1993 x3 + (1993 + n) x2 - 11x + n.
Comparing coefficients: (1) a + c = 1993; (2) ac + b + d = 1993 + n; (3) bc + ad = 11; (4) bd = n. Note first that (1) implies c is integral, and then (2) implies d is integral. Subtracting (1) and (4) from (2) gives ac - a - c = bd - b - d. It is convenient to put A = a - 1, B = b - 1, C = c - 1, D = d - 1. Then we get A + C = 1991, AC = BD and (B + 1)(C + 1) + (A + 1)(D + 1) = 11. Expanding the last equation and multiplying it through by B, we get: B2(C + 1) + BC + B + ABD + BD + AB + B = 11B. Substituing AC for BD, we get B2(C + 1) + B(A + C - 9) + A2C + AC, or B2(C + 1) + 2.991B + A(A+1)C = 0. This quadratic for B must have real roots, so we require A(A+1)C(C+1) <= 9912. But A + C = 1991 and A and C are integral, so AC > 991 unless A or C = 0. Similarly, (A + 1)(C + 1) > 991 unless A or C = -1.
So (A, C) = (0, 1991), (-1, 1992), (1991, 0), or (1992, -1). But (B + 1(C + 1) + (A + 1)(D + 1) = 11, so we cannot have (A, C) = (-1, 1992) or (1992, -1) which imply 1993 divides 11. Hence AC = 0 and so BD = n = 0. Hence, (A, B, C, D) = (0, 0, 1991, -1982) or (1991, -1982, 0, 0). Thus the two integer roots are roots of x2 - x + 1 = 0 or of x2 - 1991x - 1982 = 0. But neither equation has integer roots.
An alternative solution by Nizameddin Ordulu and Ali Adalý is as follows:
-n = (x4 - 1993x3 + 1993x2 - 11x)/(x2 + 1) = (x2 - 1993x + 1992) + (1982x - 1992)/(x2 + 1). So if x is an integer, then x2 + 1 must divide (1982x - 1992) = 2(991x - 996) and hence also 2(991x - 996)(991x + 996) = 2.9912(x2 + 1) - 2(9912 + 9962). So x2 + 1 divides 2(9912 + 9962).
If you have access to Maple or Mathematica or something similar, then it is easy. You just note that 9912 + 9962 = 593.3329, where 593 and 3329 are primes. So there are x2 + 1 = 1, 2, 593, 2.593, 3329, 2.3329, 593.3329 or 2.593.3329. Hence x2 = 0, 1, 592, 2.593 - 1, 3328, 2.3329 - 1, 593.3329 - 1 or 2.593.3329 - 1. Again with the help of Maple, 592 = 2437, 2.593 - 1 = 3.5.79, 3328 = 2813, 2.3329 - 1 = 3.7.317, 593.3329 - 1 = 243213709, 2.593.3329 - 1 = 107.36899, none of which are squares. So the only possibilities are x = 0, ±1.
If x = 0, then n = 0. If x = 1, then n = 5, if x = -1, then n = -1999. So there is one integral root for n = 0, 5, -1999 and none for other values of n.
However, it is not really practical to carry out the factorisations by hand in a reasonable time. The major obstacle is getting 9912 + 9962 = 593.3329. Not many people are familiar with all 108 primes up to 593. Even if they are, testing for division by all of them is time-consuming. Checking that 2, 593, 2.593 etc are not squares is much easier.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002