STEP I 1997 question 7 solution

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Fortunately, this question only asks us to find constants, it doesn't ask us to prove they are the only solution. If we guess that the answer will be in some sense symmetrical, this is about the shortest STEP question ever!

Consider what happens when P(x) = 1 (constant polynomial).

Then . Now I don't know about you, but at this point I'm thinking, surely we're going to have . So let's assume that and hope for the best.

Consider P(x) = x. Then . But we've assumed and so we find .

Consider P(x) = x^2. Then

So if and , we get and so we put .

So we've found our coefficients, and we know they work for P(x) = 1, x, x^2.

Consider P(x) = x^3. The integral is odd so = 0, and our formula gives us u1^3+u2^3 = 0 as u1 = - u2.

So our formula works for 1,x,x^2,x^3 and so works for all cubic polynomials by linearity.

Solution by DFranklin