Lemma: Let for some prime $p$ and natural $t$ is true, that $p|2^{2^t}+1$ then $p>t$. Let $k$ -minimal integer, that $p|2^k-1$ so $k|p-1$ Then $k|2^{t+1}$ so $k=2^r,r \leq t+1$ If $r \leq t$ then $2^r|2^t \to p|2^{2^r}-1|2^{2^t}-1\to p|2^{2^t}-1 \to p=2$ -contradiction. So $r=t+1,k=2^{t+1}$ so $p\geq k+1 =2^{t+1}+1> t$ Let $n \geq m>1$ and $p|m$ - prime divisor . $p \leq m \leq n$ But from lemma follows, that $p>m$ or $p>n$ - contradiction. So $m=1$ $n|5(2^{2^n}+1)$ $n=1$ - obvious solution, let $n>1$ Let $p$ - prime divisor of $n$ and $p \neq 5$ then by lemma $p>n$ - contradiction, so $p=5$ and $n=5^r$ But for $n\geq 2$ true that $2^{2^n}+1=16^{2^{n-2}}+1=2 \pmod 5$ so $25\not | 5(2^{2^n}+1)$ so $n=5$ Answer: $\boxed{(m,n)=(1,1),(1,5),(5,1)}$

fastlikearabbit wrote: Find all positive integers $m$ and $n$ such that $(2^{2^{n}}+1)(2^{2^{m}}+1) $ is divisible by $m\cdot n $ . see here Bulgaria 2016 problem 1

Let us first assume that $m,n>1$ and $F_x$ the $x$ -th Fermat number. Lemma: Let $n$ be a positive integer. Then $gcd(n,F_n)=1$ Proof: Let $r$ be a prime such that $r|n$ and $r|F_n$. Then we easily see that $2^{m+1}|r-1$ So, $r>2^{m+1}>m \geq r$ a contradiction. So $gcd(n,F_n)=1$ From this lemma, we see that $m|2^{2^n}+1$ and $n|2^{2^m}+1$ WLOG, say $m>n$. By Édouard Lucas, we know that any prime divisor of $F_n$ is of the form $k \cdot 2^{n+2}+1$ So $n \geq 2^{m+2}+1 > m \geq 2^{n+2}+1 > n$ an apparent contradiction. So we get that at least one of the numbers must be equal to $1$. And putting $n=1$, we get $m=5$ or $m=1$. so all solutions are $(m,n) = (1,1), (1,5), (5,1)$. QED

order of element

fastlikearabbit wrote: Find all positive integers $m$ and $n$ such that $(2^{2^{n}}+1)(2^{2^{m}}+1) $ is divisible by $m\cdot n $ . Taking orders we have that if $p|2^{2^n}+1$ then $p=1(mod 2^{n+1})$ So if $m,n$ diferent from $1$ we must have $(2^{n+1}+1)(2^{m+1}+1)<=mn$ contradiction. And now we have the couples $(m,n)=(1,1),(5,1)$

Solved with dolphinday. Another problem which uses 'Krishna Lemma'. We claim that the answers are \[(n,m)=(1,1),(1,5) \text{ and }(5,1)\] We observe immediately that $m$ and $n$ must be odd. Further, WLOG assume that $n \geq m$. If $m>1$ then there exists a prime $p \mid m$. Now, \[p \mid 2^{2^m} +1 \text{ or } p \mid 2^{2^n}+1\] Now comes the famous lemma. Claim (Krishna Lemma) : For all primes $p$ and positive integers $k$, if $p \mid 2^{2^k}+1$ then, $p\equiv 1 \pmod{2^{2^{k+1}}}$. Proof : Note that, \[2^{2^k} \equiv -1 \pmod{p} \implies 2^{2^{k+1}} \equiv 1 \pmod{p}\]Thus, $\text{ord}_p(2) \mid 2^{k+1} $ but $\text{ord}_p(2) \nmid 2^{2^{k}}$. This implies that $\text{ord}_p(2)=2^{k+1}$. But, we know that by Fermat's Little Theorem, $\text{ord}_p(2) \mid p-1$ which implies that, $2^{2^{k+1}} \mid p-1$. Thus, \[p \equiv 1 \pmod{2^{2^{k+1}}}\]as claimed. Now, by applying Krishna Lemma to the above result, we have that $p\equiv 1 \pmod{2^{2^{m+1}}}$ or $p\equiv 1 \pmod{2^{2^{n+1}}}$. Since $p>1$ this implies that, \[p\geq 2^{2^{m+1}}+1\]But since $p \mid m$, we also know that $m \geq p$. Thus, \[m \geq 2^{2^{m+1}}+1\]which is false for all positive integers $m > 1$. But this is a clear contradiction, so we conclude that $m=1$. Plugging this in we are left with \[n \mid 5\left(2^{2^n}+1\right)\]Now, if there exists a prime $p\neq 5$ such that $p \mid n$, as before, we obtain a size contradiction. Thus, $n=1$ or $n=5^k$ for some positive integer $k$. Further, \[2^{2^n}+1 \equiv 1+1 \equiv 2 \pmod{5}\]for all $n>1$. Thus, $5 \nmid 2^{2^n}+1$ which implies that \[\nu_5(n) \leq \nu_5(5(2^{2^n}+1)) = \nu_5(5)+\nu_5(5(2^{2^n}+1))= 1\]This means, $k\leq 1$. So $n=1$ or $n=5$ which can both be seen to work upon substitution. We are done.

We will denote order of 2 mod m as $\delta_m(2)$. Lemma: For $x'|x$, if $x'|2^{2^x}+1$ then $x'=1$. Proof:Suppose $x' \neq 1$, then note that $x'|2^{2^x}+1 \implies 2 \nmid x'$. Thus $\delta_{x'}(2) \nmid 2^x, \delta_{x'}(2) \mid 2^{x+1} \implies \delta_{x'}(2)=2^{x+1}$. Thus $2^{x+1} \mid \varphi(x')$ So $x\geq x' \geq \varphi(x') \geq \nu_2(\varphi(x')) > x+1$ which is absurd. Thus $x'=1$. This means we can only have two cases for each number, $m=1$ or $m|2^{2^n}+1$ and vice versa for $n$. First note that under no conditions can $2|m,n$. If $m=1$, $n|2^{2^m}+1$ then $n|5 \implies n=1,5$. If $m=1$, $n=1$ then $(m,n)=(1,1)$. If $m|2^{2^n}+1, n|2^{2^m}+1$ then we get that if $m,n \neq 1$, $\delta_{m}(2)=2^{n+1} \implies 2^{n+1} \mid \varphi(m) \implies n < 2^{n+1} \leq \varphi(m) \leq m$. Similarly for $n$ we find that $m<2^{m+1} \leq \varphi(n) \leq n$ which means $n<m, m<n$ which is absurd. If one of $n,m=1$ then we can see that this yields the same sets of solutions we have found already. Thus the solutions are $(1,1),(1,5),(5,1)$.