| Solution by Christopher J. Bradley, Clifton College, Bristol, U.K.
Suppose there exists a positive number k such that
x[1]^2 + x[2]^2 + ... + x[n]^2 = k^2 and
x[1]^3 + x[2]^3 + ... + x[n]^3 = k^3
thereby providing a possible solution; then
(x[1]-k)*x[1]^2 + (x[2]-k)*x[2]^2 + ... + (x[n]-k)*x[n]^2 = 0.
Now either all the terms are zero or one of the terms at least is
positive. Suppose (x[i]-k)*x[i]^2 > 0; then x[i] > k, contradicting
sum(x[i]^2) = k^2. Hence all solutions arise from when the terms are
all zero, and are thus of the form
x[j] = k > 0 and x[i] = 0 for all i <> j
for some j in { 1, 2, ..., n }, or else all x[i] = 0. [Note that from
sum(x[i]^2) = k^2, at most one x[i] can equal k. -- Ed.]
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