3rd Putnam 1940

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Problem A6

p(x) is a polynomial with real coefficients and derivative r(x) = p'(x). For some positive integers a, b, ra(x) divides pb(x). Prove that for some real numbers A and α and for some integer n, we have p(x) = A(x - α)n.

 

Solution

Straightforward.

Write p(x) = A ∏ (x - αi)ni.
Then r(x) = p(x) ∑ ni/(x - αi) = (A ∏ (x - αi)ni-1 ) q(x), where no αi is a root of q(x). This is easily seen, because q(x) is a sum of terms, all but one of which has a factor (x - αi). But q(x) divides p(x)b which has no roots except the αi. Hence q(x) must be a constant. But now the degree of r(x) is wrong unless there is just one αi.

 


 

3rd Putnam 1940

© John Scholes
jscholes@kalva.demon.co.uk
15 Sep 1999