Zsigmondy yields $n=2$

If $p\equiv 1 \pmod 3$ then $p^n-(p-1)^n \equiv 1 \pmod 3$ So $p \equiv 2 \pmod 3$ If $d|n$ is divisor of $n$ than $p^d-(p-1)^d$ is a power of $3$ too, so $3|2^d-1 \to d$ is even. So $n=2^k$ If $k\geq 2$ $p^2-(p-1)^2 |p^{n}-(p-1)^{n}$ so $2p-1=3^a$ $p^2+(p-1)^2|p^{n}-(p-1)^{n}$ so $2p^2-2p+1=3^b \to 2*3^b=4p^2-4p+2=3^{2a}+1 \to min(2a,b)=0 \to a=0 \to p=2 \to 5=3^b \to$ contradiction. So $n=2$

$p^n-(p-1)^n=3^a$ $p=3$ yields no solution. If $p \equiv 1 \pmod 3$, then $LHS \equiv 1 \pmod 3$ If $p \equiv 2 \pmod 3$ then $LHS \equiv 2^n-1 \pmod 3 \implies n=2k$ $[p^k-(p-1)^k][p^k+(p-1)^k]=3^a$ $p^k$ and $(p-1)^k$ are coprime. Let $gcd (p^k+ (p-1)^k, p^k- (p-1)^k)= d$ $d|p^k+ (p-1)^k$ and $d|p^k-(p-1)^k$ $ \implies$ $d|p^k+(p-1)^k+p^k-(p-1)^k$ $ \implies d|2p^k$. Similarly we can show that $d|2(p-1)^k$. Hence $d$ is either $1$ or $2$. $p^k+ (p-1)^k$ and $p^k- (p-1)^k$ are both odd so $d=1$. The only possibility is: $p^k-(p-1)^k=1$ $(1)$ and $p^k+(p-1)^k=3^a$ Applying Mihailescu's theorem on $(1)$, or either using simple inequalities we conclude that $k=1 \implies n=2$. $p=2, 5 ...$ statify the equation for $n=2$

We have $n>1$ since $3^0$ is not considered a power of $3$. Note that $p\equiv 2\pmod{3}$, or else one of $p^n$ or $p^{n-1}$ would be divisible by $3$, and the other not. Thus, we have \[3\mid p^n-p^{n-1}\implies 2^n\equiv 1\pmod{3},\]so $n$ is even. If $4\mid n$, then we have that \[p^4-(p-1)^4=(p^2-(p-1)^2)(p^2+(p-1)^2)\]is a power of $3$. We have $p\equiv 2\pmod{3}$, so $p^2+(p-1)^2\equiv 2\pmod{3}$, which can't happen. Thus, $n=2k$ for some odd $k$. But then $p^k-(p-1)^k\mid p^n-(p-1)^n$, so $p^k-(p-1)^k$ is a power of $3$. By our previous work thus can't happen unless $k=1$, so the only possibility is $\boxed{n=2}$. This works as $5^2-4^2=3^2$. Alternatively, Zsigmondy just nukes the problem, but that's boring.

This one was proposed by me.

yayups wrote: Alternatively, Zsigmondy just nukes the problem, but that's boring. Could you please show me how to use Zsigmondy here? I've never used it in a problem before. Thanks

Vertil wrote: yayups wrote: Alternatively, Zsigmondy just nukes the problem, but that's boring. Could you please show me how to use Zsigmondy here? I've never used it in a problem before. Thanks OK. In case $n$ is odd it is straightforward to show that $3$ cannot divide $p^n-(p-1)^n$. So $n$ must be even. Let $n=2k$. $3|p^2-(p-1)^2$.Thus if $k>1$ $p^{2k}-(p-1)^{2k}$ should have a prime divisor other than three, which is nonsense. Hense. $n=2$ etc

Thanks, Hamel!

yayups wrote: We have $n>1$ since $3^0$ is not considered a power of $3$. Note that $p\equiv 2\pmod{3}$, or else one of $p^n$ or $p^{n-1}$ would be divisible by $3$, and the other not. Thus, we have \[3\mid p^n-p^{n-1}\implies 2^n\equiv 1\pmod{3},\]so $n$ is even. If $4\mid n$, then we have that \[p^4-(p-1)^4=(p^2-(p-1)^2)(p^2+(p-1)^2)\]is a power of $3$. We have $p\equiv 2\pmod{3}$, so $p^2+(p-1)^2\equiv 2\pmod{3}$, which can't happen. Thus, $n=2k$ for some odd $k$. But then $p^k-(p-1)^k\mid p^n-(p-1)^n$, so $p^k-(p-1)^k$ is a power of $3$. By our previous work thus can't happen unless $k=1$, so the only possibility is $\boxed{n=2}$. This works as $5^2-4^2=3^2$. Alternatively, Zsigmondy just nukes the problem, but that's boring. sorry i did not understand why it can't happen unless $k=1$ ?

Supermathlet_04 wrote: $p^n-(p-1)^n=3^a$ $p=3$ yields no solution. If $p \equiv 1 \pmod 3$, then $LHS \equiv 1 \pmod 3$ If $p \equiv 2 \pmod 3$ then $LHS \equiv 2^n-1 \pmod 3 \implies n=2k$ $[p^k-(p-1)^k][p^k+(p-1)^k]=3^a$ $p^k$ and $(p-1)^k$ are coprime. Let $gcd (p^k+ (p-1)^k, p^k- (p-1)^k)= d$ $d|p^k+ (p-1)^k$ and $d|p^k-(p-1)^k$ $ \implies$ $d|p^k+(p-1)^k+p^k-(p-1)^k$ $ \implies d|2p^k$. Similarly we can show that $d|2(p-1)^k$. Hence $d$ is either $1$ or $2$. $p^k+ (p-1)^k$ and $p^k- (p-1)^k$ are both odd so $d=1$. The only possibility is: $p^k-(p-1)^k=1$ $(1)$ and $p^k+(p-1)^k=3^a$ Applying Mihailescu's theorem on $(1)$, or either using simple inequalities we conclude that $k=1 \implies n=2$. $p=2, 5 ...$ statify the equation for $n=2$

Clearly for $p\equiv 1\mod 3$ we have $p^n-(p-1)^n\equiv 1\mod 3$ ,So $p\equiv 2\mod 3$ for odd $n$ we have $p^n\equiv 2\mod 3,(p-1)^n\equiv 1\mod 3$ hence for odd $n$ $p^n-(p-1)^n\equiv 1\mod 3$ for $n=2x$ we have $p^n\equiv 1\mod 3,(p-1)^n\equiv 1\mod 3$ hence for $n=2x$ $p^n-(p-1)^n\equiv 0\mod 3$ hence $p\equiv 2\mod 3,n=2x$ is only possible. So $(p^x-(p-1)^x)(p^x+(p-1)^x)=3^a$ Case 1-: for $x=1$ we get $2p-1=3^a\implies 2p=3^a+1$ which has a solutions such as for odd $a$ there is only 1 solution at $p=2$ , for even $a$ solutions are $(a=2, p=5), (a=4, p=41)..$ So there are solutions for $n=2$ case. Case 2-: $x>1$ $p^x-(p-1)^x=3^b, p^x+(p-1)^x=3^{a-b}$ then by applying same logic as above we get $x=2y , p\equiv 2\mod 3$ but then $p^x+(p-1)^x\equiv 2\mod 3$ which is contradiction. So $p^x-(p-1)^x =1$ which has no solution by Fermat's last Theorem Hence $n=2$ is Only possible $\blacksquare$