Problem: Factor .
Solution: Consider the polynomial . We have
so we want to factor the second term when . Call it so that . Consider the relation
Since is a root of the LHS, we factor it out of the RHS as well to get
Dividing through by and rearranging, we obtain the nice expression
Letting , this becomes
which is conveniently enough a difference of two squares. And we’ll leave it as this because the factors are not particularly nice or anything. QED.
Comment: This “natural” factorization was basically the one crucial step to solving the USAMO #5 this year and most people did not see it, unsurprisingly. Taking the difference was the trickiest/cleverest part, and there were definitely a limited number of approaches to this factorization. Oh well, at least it seems sort of cool after you know about it.
Practice Problem: (2007 USAMO – #5) Prove that for every nonnegative integer n, the number is the product of at least (not necessarily distinct) primes.
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