Find all real numbers a for which there exist non-negative real numbers x1, x2, x3, x4, x5 satisfying:
x1 + 2x2 + 3x3 + 4x4 + 5x5 = a,
x1 + 23x2 + 33x3 + 43x4 + 53x5 = a2,
x1 + 25x2 + 35x3 + 45x4 + 55x5 = a3.
Solution
Take a2 x 1st equ - 2a x 2nd equ + 3rd equ. The rhs is 0. On the lhs the coefficient of xn is a2n - 2an3 + n5 = n(a - n2)2. So the lhs is a sum of non-negative terms. Hence each term must be zero separately, so for each n either xn = 0 or a = n2. So there are just 5 solutions, corresponding to a = 1, 4, 9, 16, 25. We can check that each of these gives a solution. [For a = n2, xn = n and the other xi are zero.]
Solutions are also available in Murray S Klamkin, International Mathematical Olympiads 1978-1985, MAA 1986, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
12 Oct 1998