If $ p = 2$ then $ (p - 1)^{n} + 1 = 2$ so $ n = 2$. If $ n = 1$ then $ (p - 1)^{n} + 1$ is divisible by $ n^{p - 1}$ for any prime $ p$. If $ n = 2$, then $ (p - 1)^{n} + 1$ must be even so $ (p - 1)^{n}$ is odd so $ p$ is even but it is prime therefore $ p = 2$. This gives the solutions $ (1,p)$ and $ 2,2$ when $ n\leq 3$ or $ p\leq 3$. Now look at the case when $ p,n\geq 3$. Because then $ p$ is odd, $ (p - 1)^{n} + 1$ is odd so $ n$ is odd. Let $ x$ be the greatest prime divisor of $ n$, then we have $ x|n|n^{p - 1}|((p - 1)^{n} + 1)$ so $ (p - 1)^{n}\equiv - 1\pmod x$. Now because $ x|n$ then $ x - 1$ and $ n$ are coprime so there will exist positive integers $ a,b$ such that $ an = b(x - 1) + 1$. Because $ n$ is odd, $ x$ is odd so $ b(x - 1)$ is odd so $ b(x - 1) + 1$ is even hence $ a$ is odd. But then, if $ (p - 1)^{n}\equiv - 1\pmod x$, $ (p - 1)^{an}\equiv - 1\pmod x$ therefore $ (p - 1)^{b(x - 1)}*(p - 1)\equiv - 1\pmod x$. But by Fermat's Little Theorem $ (p - 1)^{x - 1}\equiv 1\pmod x$ because $ x$ and $ p - 1$ are coprime (otherwise, it would be impossible to have $ n^{p - 1}|((p - 1)^{n} + 1)$ as $ x|n$. Then, $ (p - 1)^{b(x - 1)}*(p - 1)\equiv - 1\pmod x$ gives $ (1)^{b}*(p - 1)\equiv - 1\pmod x$ so $ p - 1\equiv - 1\pmod x$ so $ x|p$ so $ x = p$. But $ x|n$ and $ n < 2p$ so $ n = p$. Then $ n^{p - 1}|((p - 1)^{n} + 1)$ is equivalent to $ p^{p - 1 }|((p - 1)^{p} + 1)$ but if we expand $ (p - 1)^{p} + 1$ we get: $ p^{p - 1}| p^{p} - p^{p - 1}*\binom{p}{1} + p^{p - 2}*\binom{p}{2} + ... + p*\binom{p}{1} - 1 + 1$ or: $ p^{p - 1}|p^{2}(p^{p - 2} - p^{p - 3}*\binom{p}{1} + ... + \binom{p}{p - 2} + 1)$ and therefore $ p^{p - 1}|p^{2}$ because the expression in brackets is not divisible by $ p$ hence the only solution possible is if $ p - 1\leq 2$ so $ p\leq 3$. But we assumed $ p\geq 3$ so $ p = 3$ so $ n = 3$. So the solutions are $ (1,p), (2,2), (3,3)$

If not have n < 2p , can you prove it ?

It is an IMO problem and it also relate with an IMO problem Find all positive integer n such that :$\frac {2^n+1}{n^2}$ is an integer $(*)$ (put p=3 in A72)... DR.TITU ANDREESCU wrote a comment in his book(Number theory:structures,examples and problems),the condition $n<2p$ is not necessary. Very lucky that i found a solution of $(*)$ difference from the official solution by Lifting the exponent lemma in this link http://reflections.awesomemath.org/2007_3.html Work similarly,i think we can prove $A72$ without condition $n<2p$

taylorcute56 wrote: If not have n < 2p , can you prove it ? Suppose $ p>2 $ and $ n>1 $. (otherwise, the case is trivial.) Since $p$ is odd prime, $n>1$ is odd integer. Suppose $q$ is the samllest prime divisor of $n$ s.t. $ q|n|n^{p-1}|(p-1)^{n}+1 $. $ (p-1)^{n}\equiv-1\pmod q $.................................. (1) $ (p-1)^{q-1}\equiv 1\pmod q $ $ (p-1)^{gcd(2n, q-1)}\equiv 1\pmod q $ So, $ (p-1)^{2}\equiv 1\pmod q $ So, $ p\equiv 2\pmod q $ or $ q|p $ If $ p\equiv 2\pmod q $ , by (1) and $q>2$, contraduction. So, $ q|p $ So, $q=p$ So, $ p|n $ $n=p^{s}u$, $ s\geq 1 $, $ p>2 $, $(u,2)=(u,p)=1$ $ n^{p-1}|(p-1)^{p^{s}u}+1 $ $ (p-1)^{p^{s}u}+1 \equiv p^{s+1} ( \mod p^{s+2})$ $(p-1)s =v_{p}( n^{p-1}) \leq v_{p}( (p-1)^{p^{s}u}+1) = s+1 $ $(p-1)s \leq s+1$ $ 2s+1 \geq sp $ $ 3 \geq 2+ \frac{1}{s} \geq p \geq 3 $ so, $s=1$ and $p=3$. so, $n=3u$, $ u\geq 1 $, $(u,2)=(u,3)=1$. Suppose $u>1$. Suppose $r$ is the samllest prime divisor of $u$ s.t. $ r|u|9u^{2}|8^{u}+1 $. So, $r>3$. $ 8^{u}\equiv-1\pmod r $ $ 8^{r-1}\equiv 1\pmod q $ $ 8^{gcd(2u, r-1)}\equiv 1\pmod r $ $ 8^{2}\equiv 1\pmod r $ $ r|63 $ So, $r=7$ So, $ 7 | 8^{u}+1 \equiv 2\pmod 7$, contradiction. So, $u=1$ So, $n=3$ So, $(n,p)=(3,3)$. .

Thanh.Minh2009 wrote: Find all positive integer n such that :$\frac {2^n+1}{n^2}$ is an integer $(*)$ Suppose $n>1$ First, $ n|2^{n}+1 $ Suppose $p>2$ is the smallset prime divisor of$n$. $ 2^{n}\equiv-1\pmod p $ $ 2^{p-1}\equiv 1\pmod p $ $ 2^{gcd(2n,p-1)}\equiv 1\pmod p $ So, $p=3$ $n=3^{s}u$, $s\geq 1$, $(u,2)=(u,3)=1$ $ 3^{2s}u^{2}|2^{3^{s}u}+1 $ $2^{u}=x\equiv-1\pmod 3$ $ x^{3^{s}}+1 =(x+1)\prod_{i=0}^{s-1}(x^{3^{i}2}-x^{3^{i}}+1) $ For $i=0, 1, ..., s-1$ $x^{3^{i}2}-x^{3^{i}}+1=(x^{3^{i}}+1)(x^{3^{i}}-2)+3 \equiv 3\pmod 9 $ and since $(u,2)=(u,3)=1$, $x+1=2^{u}+1\equiv -3, 3\pmod 9 $ So, $ x^{3^{s}}+1$ have exactly $(s+1)$ $3$'s as prime divisor. So, $2s\leq s+1$ So, $s\leq1$ So, $s=1$. So, $n=3u$, $ u\geq 1 $, $(u,2)=(u,3)=1$. So, $ 9u^{2}|8^{u}+1 $. As we have said in previous post, luapsedree wrote: Suppose $u>1$. Suppose $r$ is the samllest prime divisor of $u$ s.t. $ r|u|9u^{2}|8^{u}+1 $. So, $r>3$. $ 8^{u}\equiv-1\pmod r $ $ 8^{r-1}\equiv 1\pmod q $ $ 8^{gcd(2u, r-1)}\equiv 1\pmod r $ $ 8^{2}\equiv 1\pmod r $ $ r|63 $ So, $r=7$ So, $ 7 | 8^{u}+1 \equiv 2\pmod 7$, contradiction. So, $u=1$ So, $n=3$ So, $n=3$. .

Let $q$ be the smallest prime divisor of $n$. Hence $q|(p-1)^{n}-1$. Again from F.L.T we get $q|(p-1)^{q-1}-1$. Hence we get that $q|(p-1)^{gcd(q-1,2n)}-1$. Since q is the smallest prime divisor of $n$ so $gcd(2n,q-1)=2$. Hence $q|(p-1)^{2}-1$ i.e $q|p$ or $q|p-2$. Since $gcd((p-1)^{n}-1,p-2)=1$ so we conclude that $q|p$. and so $q=p$. Hence $p|n$. Now $n<2p$.So $n=p$. Now the problem is to find all primes such that $p^{p-1}|(p-1)^{p}-1$. Now let $p>4$. Then $p^{3}|(p-1)^{p}-1$ which forces that $p^{3}|p$. a contradiction. so $p<4$ and hence $p=2,3$.Now we check that these two values satisfies the required relations and hence $(n,p)=(2,2),(3,3)$.

rem wrote: Now because $ x|n$ then $ x - 1$ and $ n$ are coprime Is this always true?? What about $n=21$, and $x=7$, for example??