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I encountered this video about number theory question from Swedish Maths Olympiad:

Find all integers $n\geq 8$ such that $n^{\frac{1}{n-7}}$ is also an integer.

The video shows step by step solution. However, I think the question can be solved much easier as the following:

We substitute $x=n-7$ and $n=a^{x}$ to obtain $a^{x}=x+7$. Then quite obviously we find that for $x\geq4$ we have $a<2$ and we only need to evaluate $x=1,2,3$.

Then we find that the only pairs $(a,x)$ are $(8,1)$ and $(3,2)$ i.e. the only solutions are $n=8,9$.

Here is the link of the video: Swedish Mathematics Olympiad

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    $\begingroup$ +1 looks flawless to me. $\endgroup$ Commented Sep 30, 2020 at 22:08
  • $\begingroup$ 9/(9 - 7) is not an integer. $\endgroup$ Commented Oct 1, 2020 at 1:29
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    $\begingroup$ @WilliamElliot do you mean $9^{\frac{1}{9-7}}=3$? $\endgroup$
    – acat3
    Commented Oct 1, 2020 at 1:32
  • $\begingroup$ You forgot to formulate explicitely your question. $\endgroup$ Commented Jul 23, 2023 at 13:31

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