Putnam 1998

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Problem B6

Show that for any integers a, b, c we can find a positive integer n such that n3 + a n2 + b n + c is not a perfect square.

 

Solution

Easy. Almost any approach works.

In fact, f(n) = n3 + a n2 + b n + c cannot be a perfect square for all of n = 1, 2, 3 and 4. Perfect squares must be = 0 or 1 (mod 4). But f(2) -f(4) = 2b (mod 4), so if f(2) and f(4) are perfect squares then 2b = 0 (mod 4), and f(3) - f(1) = 2b + 2 (mod 4), so if f(1) and f(3) are perfect squares then 2b + 2 also = 0 (mod 4), which is impossible.

 


 

Putnam 1998

© John Scholes
jscholes@kalva.demon.co.uk
12 Dec 1998