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Problem 871

AIME late

Number theory Difficulty 5.7 Prove it

9.5. Prove that any number of the form $n^{k}$ , where $n$ and $k$ are natural numbers different from 1, can be represented as the sum of $n$ consecutive odd numbers.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

9.5. For the number $a+(a+2)+\ldots+(a+2 n-2)=n(a+n-1)$ to equal $n^{k}$ , we need to set $a+n-1=n^{k-1}$ , i.e., $a=n^{k-1}-n+1$ . It is clear that the number $a$ is odd in this case.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.