TMIMO 2009, Grade 10, Solution to Problem 3

Find all natural numbers M for which (7^M - M²)^M - M^2M = 2009

Solution by Sadik Shahidain, Cranbury, NJ, USA.

Rewriting the left side as

(7^M - M²)^M - (M²)^M = 2009

Factoring the left

(7^M - 2M²)((7^M - M²)^M-1 + M(7^M - M²)^M-2(M²) + ... + M(7^M - M²)(M²)^M-2 + (M²)^M-1) = 2009

Therefore 7^M - 2M² must be a factor of 2009, and the possibilities are

7^M - 2M² = {1 or 7 or 41 or 49 or 287 or 2009}

and M < 4, where M is a natural number. 2 is the only viable option and substituting back into the original equation gives 2 as the answer.

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