A Contest Problem l

Number Theory Level 3

Let $n$ be a positive integer such that $n\ge 3$ , then

$\large n^{n^{n^{n}}}-n^{n^{n}}$

is divisible by

1989 1990 1987 1996 1988

1 solution

Hana Wehbi
Jun 1, 2019

I am not sure if I can follow the proof. in case 1, where you assumed $3 \nmid n$ and used Euler-Fermat afterwards, I do not think you have used the theorem correctly. In such a case, you can have $n^{n^{n^n}} \equiv n^{n^n} \ mod 3^2$ if $n^{n^n} \equiv n^n \ mod \phi(3^2)$ . I might be wrong, so advise me accordingly.

A Former Brilliant Member - 2 years ago

so we assume that if 3 does not divide n, then gcd(3,n)=1, then we apply Euler-Fermat theorem.

Hana Wehbi - 2 years ago

A Contest Problem l

1 solution

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