IMO 1979 SL 15

Origin: ISR

Problem

The nonnegative real numbers x1, x2, x3, x4, x5, a satisfy the following relations:  i=1 ixi = a,  i=1 i3xi = a2,  i=1 i5xi = a3. What are the possible values of a?

Solution

We note that  i=1 i(a−i2)2xi = a2  i=1 ixi−2a  i=1 i3xi+  i=1 i5xi = a2\cdota−2a\cdota2+a3 = 0. Since the terms in the sum on the left are all nonnegative, it follows that all the terms have to be 0. Thus, either xi = 0 for all i, in which case a = 0, or a = j2 for some j and xi = 0 for i ̸= j. In this case, xj = a/j = j. Hence, the only possible values of a are {0, 1, 4, 9, 16, 25}.