Hint: Lifting exponent lemma

Lifting exponent lemma solves immediately, but we can solve without using it. \[ \begin{array}{rcccccccccccccccccccc} \mbox{exponent :}&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20 \\ \mbox{powers of 2 :}&2&4&8&16&7&14&3&6&12&24&23&21&17&9&18&11&22&19&13&1 \\ \mbox{powers of 3 :}&3&9&2&6&18&4&12&11&8&24&22&16&23&19&7&21&13&14&17&1 \\ \mbox{sum :}&5&13&10&22&0&18&15&17&20&23&20&12&15&3&0&7&10&8&5&2 \end{array} \] Above figure shows the residues of $2^n$ and $3^n$ modulo $25.$ So, $25 \mid 2^n+3^n \: \Longleftrightarrow \: n \equiv 5 \pmod{20}$ or $n \equiv 15 \pmod {20}.$ $p=2 \: \Longrightarrow \: 2^p+3^p=2^2+3^2=13$ does not satisfy. $p=5 \: \Longrightarrow \: 2^p+3^p=2^5+3^5=275=5^2 \cdot 11$ does not satisfy. $p \neq 2 \: \wedge \: p \neq 5 \: \Longrightarrow \: 5 \mid 2^p+3^p$ since $p$ is odd and $25 \nmid 2^p+3^p$ since $5 \nmid p.$ So, $n>1$ gives a contradiction. Hence, we are done.

Though posted before,here is a solution $p=2,2^2+3^2=13$ If $p$ odd,$5|2^p+3^p$ but $v_5(2^p+3^p)=2$ if $p=5,$ but then $2^5+3^5=275\neq a^n,n>1$

The case $p=2$ is trivial. So now we have $p\ge3$ and therefore $p$ is odd. Let $q^k||a$ where $q$ is prime divisor of $a$. Then we have $q^{kn}||2^p-(-3)^p$. If $q=p$ we obtain $p=5$ which doesn't lead to a solution. So $q\not =p \Rightarrow gcd(p,q)=1$ and therefore by Lifting the exponent lemma we obtain $q^{kn}||2-(-3)=5$ - absurd.

my solution: Obviously when $p=2,5$ there isn't a solution so assume that $p$ is odd and $p\neq 5$ because $5\mid 2^p+3^p$ we deduce that $5\mid a^n\longrightarrow 5\mid a\longrightarrow 5^n\mid a^n\longrightarrow 5^n\mid 2^p+3^p$ but from lifting exponent lemma $e_5(2^p+3^p)=1$ which is absurd. DONE

I know this has probably been posted but I don't care cuz Im so proud that I can use lte! Lemma(LTE): For $p$ being an odd prime relatively to $a$ and $b$ with $p|a-b$ and $n $ is an odd postive integer. Then: $$v_p(a^n+b^n)=v_p(a+b)+v_p(n)$$ Sol: Note that for p=2 , we have that $a=13, n=1$, contradiction. Now, for $p\geq 2$, we get that $v_5(2^p+3^p)=v_5(5)+v_5(p)=1++v_5(p)=2$iff p=5. However, this leads to $2^5+3^5=275$, another contradiction. Other cases follow anagolously.

WakeUp wrote: Prove that for any prime number $p$ the equation $2^p+3^p=a^n$ has no solution $(a,n)$ in integers greater than $1$. Was searching for a different application but you can say if p=2 n=1 a=13, if p>2 odd then 5\mid2^p+3^p=a^n\implies a=5k; looking at v_5 in RHS and LHS, LHS is 2 if p=5 by LTE and 1 otherwise, so n is 2 or 1; if 1 we're done if 2 this can be checked to not work (275=2^5+3^5=(5k)^2 is impossible)

Case 1: $p=2$ $a^n=13$ which has no solution for $n>1$ Case 2:$p>2$ Using LTE we get that $v_5(2^p+3^p)=1+v_5(p)$ If $p=5$ then $a^n=275$ which has no solution for $n>1$ $\implies p\neq5\implies v_5(2^p+3^p)=1\implies v_5(a^n)=1\implies n=1$ contradiction We have proven that it has no solution

HINT: Just use Zsigmondy' s theorem, bro

Would anyone help me : How to get : (a) $25 \mid 2^n+3^n?$ (b) $n \equiv 5 \pmod{20}?$ (c) $n \equiv 15 \pmod {20}?$