https://aops.com/community/p11377037 Gluncho wrote: If $n>3$, just note that for every $p|(n-1)$, $v_p(n^{n-1}-1)=2v_p(n-1)$, implying that $(n-1)^2 | (n^{n-1}-1)$, so it's not squarefree. Hence $n=1,2$ is the only answer

please, why $v_p(n^{n-1}-1)=2v_p(n-1)$ ?

bebeevan wrote: please, why $v_p(n^{n-1}-1)=2v_p(n-1)$ ? Lifting The Exponent (lemma) It's a pretty useful tool.

jasperE3 wrote: https://aops.com/community/p11377037 Gluncho wrote: If $n>3$, just note that for every $p|(n-1)$, $v_p(n^{n-1}-1)=2v_p(n-1)$, implying that $(n-1)^2 | (n^{n-1}-1)$, so it's not squarefree. Hence $n=1,2$ is the only answer I don't think this works for $v_2(n-1)=1$

If there exists p prime odd such that p divides n-1,just apply LTE,if n=2^k+1,where k is a natural number,then n^(n-1)-1 is divisible by n-1=2^k which is divisible by 4,so we are done.It remains the case k=1 or 0 so n is 2 or 3.It verifies only n=2.

motannoir wrote: If there exists $p$ prime odd such that $p$ divides $n-1$, just apply LTE, if $n=2^k+1$, where $k$ is a natural number, then $n^{n-1} - 1$ is divisible by $n-1=2^k$ which is divisible by $4$, so we are done. It remains the case $k=1$ or $0$ so $n$ is $2$ or $3$.It verifies only $n=2$.

motannoir wrote: If there exists $p$ prime odd such that $p$ divides $n-1$, just apply LTE, if $n=2^k+1$, where $k$ is a natural number, then $n^{n-1} - 1$ is divisible by $n-1=2^k$ which is divisible by $4$, so we are done. It remains the case $k=1$ or $0$ so $n$ is $2$ or $3$.It verifies only $n=2$. Thank you for clarifying.

No need for Lifting the exponent, just note that $n^{n-1} - 1 = (n-1)(n^{n-2} + n^{n-3} + \cdots + 1)$ and that each term in the second factor is congruent to $1$ mod $n-1$, thus $(n-1)^2 \mid n^{n-1} - 1$, implying $(n-1)^2 = 1$, i.e. $n=2$.

We consider integers $n \geq 2$, such that $n^{n-1}-1$ is a positive integer. Clearly when $n=2$, $n^{n-1}-1=1$ is square-free. Assume some odd prime $p$ divides $n-1$. Then $\nu_{p}(n^{n-1}-1)=2\nu_p(n-1) \geq 2$, so $n^{n-1}-1$ is not square free. Otherwise, $n-1=2^k$ for some integer $k \geq 1$, then $\nu_{2}(n^{n-1}-1)=2\nu_{2}(n-1)+\nu_{2}(n+1)-1 \geq 2k \geq 2$, so $n^{n-1}-1$ is not square free. So the only answer is $n=2$.