Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Assume $n$ is not a power of $2$. In this case there is an odd number $k\ge 3$ s.t. $2^k-1|2^n-1|m^2+9$. $2^k-1$ is of the form $4t+3$, so it has a prime factor of the form $4t+3$ and $\ne 3$, because $3\not |\ 2^k-1$, so it can't divide $m^2+9$ because if a prime of the form $4t+3$ divides $a^2+b^2$, then it divides $a,b$. This means that $n$ must be a power of $2$. On the other hand, assume $n=2^k$. In this case, the only prime of the form $4t+3$ which divides $2^n-1$ is $3$. Indeed, let $p$ be such a prime. We then have $p|(2^{2^k}-1,2^{p-1}-1)=2^{(2^k,p-1)}-1=2^2-1=3$ (because $p-1=4t+2$). At the same time, $9\not |2^n-1$ (because $9|2^n-1\iff 6|n$), so $2^n-1$ is of the form $3\alpha$, where $\alpha$ has only divisors of the form $4t+1$, so there is some $m'$ s.t. $\alpha|m'^2+1$. All we need to do now is to take $m=3m'$, and we have found $m$ s.t. $2^n-1|m^2+9$.

grobber wrote: Assume $n$ is not a power of $2$. In this case there is an odd number $k\ge 3$ s.t. $2^k-1|2^n-1|m^2+9$. $2^k-1$ is of the form $4k+3$, so it has a prime factor of the form $4t+3$ and $\ne 3$, because $3\not |\ 2^k-1$, so it can't divide $m^2+9$ because if a prime of the form $4t+3$ divides $a^2+b^2$, then it divides $a,b$. Hi here grobber has used that if an number is of the form $4k+3$ then it will a prime divisor of the same form, and also, if a number of the form $4k+3$ divides $a^2+b^2$ then it have to divide $a, b$ Abdurashid

And what do you want to tell us by this¿

i think he wants the proofs for this facts davron

Hm...my solution is much much longer and uglier than grobbers but since ML is a kind of archieve of solutions I've found (and they are precious few), I think I'll post it:

abdurashidjon wrote: grobber wrote: Assume $n$ is not a power of $2$. In this case there is an odd number $k\ge 3$ s.t. $2^k-1|2^n-1|m^2+9$. $2^k-1$ is of the form $4k+3$, so it has a prime factor of the form $4t+3$ and $\ne 3$, because $3\not |\ 2^k-1$, so it can't divide $m^2+9$ because if a prime of the form $4t+3$ divides $a^2+b^2$, then it divides $a,b$. Hi here grobber has used that if an number is of the form $4k+3$ then it will a prime divisor of the same form, and also, if a number of the form $4k+3$ divides $a^2+b^2$ then it have to divide $a, b$ Abdurashid Seven years ago, but for future reference, here is the proof. Consider a number of the form $4k+3.$ Clearly, all its prime factors must be of the form $4m+1$ or $4m+3.$ Since $4k+3 \equiv 3\pmod 4,$ the factors cannot all be $1\pmod 4,$ so there must be at least one prime factor of the form $4m+3.$ Now suppose that a prime $4m+3$ divides $a^2+b^2.$ Let $a^2 \equiv j\pmod{4m+3}.$ Assume for sake of contradiction that $j \neq 0.$ Then $b^2 \equiv -j\pmod{4m+3}.$ So we obtain $a^{4m+2} \equiv j^{2m+1}\pmod{4m+3}$ and $b^{4m+2} \equiv -j^{2m+1}\pmod{4m+3}.$ However, by FLT, we must have $j^{2m+1}=-j^{2m+1}=1,$ a contradiction. Therefore, $a \equiv b \equiv 0\pmod{4m+3}.$

I shall be using the fact that if any prime $p$ of the form $4k-1$ divides $m^2+9$ then $p=3$. Firstly as $2^n-1\equiv -1\pmod{4}$ for $n>1$ we have $n=2s$ for some integer $s$, otherwise $3\nmid 2^n-1$. If $n$ has any prime factor $q\neq 2$, then $2^q-1\nmid 2^s-1$. But $2^q-1$ has a prime factor of the form $4k-1$ which is greater than $3$. Hence $n=2^t$ for some integer $t$. The rest is easy, let $2^{2^t}-1=3p_1^{\alpha_1}\cdots p_n^{\alpha_n}$. By Chinese Remainder Theorem there exists a solution to \[\begin{cases}X\equiv x_1\pmod{p_1^{\alpha_1}}\\ \cdots\\X\equiv x_n\pmod{p_n^{\alpha_n}}\end{cases}\] Where $p_i^{\alpha_i}\mid x_i^2+1$. Easy to see such $x_i$ exists. Now $3X$ suffices. SO the answer is $\boxed{n=2^t}$ for some non-negative integer $t$.

$2^n - 1 \mid m^2 + 9$ Let's assume $n$ has a prime divisor $p$ different than $2$ Therefore $2^p-1\mid2^n-1$, and $2^p-1$ has a prime divisor which is different than $3$, and it's equivalent $3$ in $\mod4$. Assume it's $q$ Since $q\mid 2^p-1\mid m^2+9$, $-9$ is a quadratic residue in $\mod q$ By Legendre's Symbol, we have $1=\left(\dfrac{-9}{q}\right)=\left(\dfrac{-1}{q}\right) \cdot \left(\dfrac{9}{q}\right)$ Since $\left(\dfrac{9}{q}\right)$ is always equal to $1$, $\left(\dfrac{-1}{q}\right)$ must also be equal to $1$ Hence, $q \not\equiv 3 \mod4$. Contradiction. Therefore, $n$ can't have a prime divisor different than $2$. So $n=2^k$ Now, we will use induction to prove $n=2^k$ always satisfies. We will be done if $2^n-1$ doesn't have any prime divisor different than $3$, and equivalent $3$ in $\mod4$. Because if it really doesn't, $-1$ and so $-9$ will be a quadratic residue in all of it's prime divisors modulo, so it also will be in modulo $2^n-1$. And that's what we want. For $k=1$, $2^n-1=3$ satisfies the condition. Assume also $2^{2^k}-1$ satisfies. $2^{2^{k+1}}-1=\left(2^{2^k}+1\right)\left(2^{2^k}-1\right)$. We know the second factor satisfies. So we have to show $2^{2^k}+1$ also satisfies. $r\mid 2^{2^k}+1 \Longrightarrow 2^{2^k} \equiv -1 (\mod r)$. So $-1$ is a quadratic residue in $\mod r$. Hence $r$ can't be equivalent $3$ in $\mod 4$. Therefore, we're done.

Eray wrote: For $k\ge3$, $2^{2^3}-1\mid 2^{2^k}-1$. So $127\mid 2^{2^k}-1$ But $2^{2^3}-1=255$!!!

I have corrcted my solution. Thank you @TheOverlord

First of all note that if $2^n-1$ has a prime divisor $p \equiv 3\pmod{4}$ other than $3$ then $-9$ is not a quadratic residue modulo $2^n-1$,because $(-9)^{\frac{p-1}{2}}=-3^{p-1} \equiv -1\pmod{p}$ and this follows at once by Euler's criterion and composite quadratic residue law. Now if $n$ has an odd prime divisor $j$ then $2^j-1$ is a factor of $2^n-1$ so the latter has a prime divisor $p \equiv 3\pmod{4}$(other than $3$ since $j$ is odd) so our condition is not satisfied. So it remains to check for $n=2^k$ ($n=1$ case is obvious). The crucial fact is that if an odd prime $p \equiv 3\pmod{p}$ other than $3$ divides $2^{2^k}-1$ then since it also divides $2^{p-1}-1$ it must also divide $2^{\text{gcd}(2^k,p-1)}-1=2^2-1=3$.This is a contradiction. Also see that $2^{2^k}-1=(2^{2^{k-1}}-1)(2^{2^{k-1}}+1)$ inducting backwards we see that the highest power of $3$ dividing it cannot exceed the highest power of $3$ dividing $2^8-1=255$,i.e, $2$.(note that the latter factor is not divisible by $3$ for $k \ge 2$) Finally we see that all the primes of $2^{2^k}-1$ other than $3$ are $\equiv 1\pmod{4}$ hence $-9$ is a quadratic residue of each of them(one may see this by applying Euler's criterion: $(-9)^{\frac{p-1}{2}}=3^{p-1} \equiv 1\pmod{p}$).Also we may choose $l$ so that the highest power of $3$ in $2^{2^k}-1$ divides $l^2+9$.Now applying CRT we may find $m$. Thus the answer to our problem is $n=1,2^k \forall k \in \mathbb{Z^+}$

An easy proof that $n$ is a power of $2$. Let $n=2^a b$ where $a\ge 0$ and $b$ is odd and let $p$ be a prime factor of $2^b-1$, hence $p\mid 2^b-1\mid 2^n-1\mid m^2+9$ and so $m^2\equiv -3^2\pmod p$. Since $b$ is odd we have $3\nmid 2^b-1$ giving $p\ne 3$, therefore the last congruence implies that $-1$ is a quadratic residue of $p$, implying that $p\equiv 1\pmod 4$. It follows that $2^b-1\equiv 1\pmod 4$, thus $b=1$. So $n=2^a$.

That pride after solving an N5 though... We call an $n$ $good$ if it satisfies the condtion, and $bad$ otherwise. We will show that the only $good$ numbers are powers of $2$. For $n=1$, the result is trivially true. So now assume $n>1$.

$p=2$. $Proof:$ Well known and easy to prove. Lemma 2: For any $n \in N$, there exists $x \in N$ such that $x^2 \equiv -1$ $\left( mod \left( 2^{2^{n}}+1 \right) \right)$.

Now, a $good$ value of $n$ (which has $p$ as a prime divisor of $2^n-1$) satifies $m^2 \equiv -9=(-1) \cdot (3)^2 \pmod {p}$ and so if $p \ne 3$, then $(m \cdot 3^{-1})^2 \equiv -1 \pmod{p}$ and so $-1$ is a quadratic residue mod $p$, implying that $p \equiv 1 \pmod{4}$. Now, we show that there exists a prime $p \ne 3$ such that $p|2^n-1$ and $p \equiv -1 \pmod{4}$ whenever $n$ is not a power of $2$ to get a contradiction.

exist a prime factor $p$ of $2^q-1$ satisfying $q \equiv -1 \pmod{4}$. Now, if $p=3$, then $1 \equiv 2^q \equiv 2 \pmod{3}$, as $q$ is odd. But this is a contradiction. Thus, $p|2^q-1|2^n-1|m^2+9$ with $p \equiv -1 \pmod{4}$ and $p \ne 3$, a contradiction as above.

$$\left\{\begin{array}{l}m \equiv m_o \left (mod \left(2^{2^{k}}-1 \right) \right)\\m \equiv 3m' \left(mod \left(2^{2^{k}}+1 \right) \right)\end{array}\right.$$Hence, we are done. $\blacksquare$

The so called Christmas theorem states that all prime factors of $n^2+1$ are $1\pmod{4}$. We have the following solution. We claim the only solutions are $n=2^k$ for any positive integer $k$. First of all, we see that \[f(k):=2^{2^k}-1=(2^{2^{k-1}}+1)(2^{2^{k-2}}+1)\cdots(2^{2^0}+1).\]Note every term is of the form $a^2+1$ except the last term which is $3$. Therefore, \[f(k)=3p_1^{e_1}\cdots p_r^{e_r}\]where $p_1,\ldots,p_r$ are distinct primes that are $1\pmod{4}$ by the Christmas theorem. Let $g_i$ be a primitive root mod $p_i^{e_i}$. We see that if $x\equiv g_i^{p_i^{e_i-1}(p_i-1)/4}$, then $p_i^{e_i}\mid x^2+1$, so by CRT, we can pick some $x$ such that $f(k)/3\mid x^2+1$. Then, we see that $f(k)\mid (3x)^2+9$, so $n=f(k)$ works. Now, suppose $n=2^k\ell$ where $\ell>1$ is odd. Note that all prime factors of $m^2+9$ are either $1\pmod{4}$ or are $3$. We see that \[2^n-1 = (2^{2^{k-1}\ell}+1)(2^{2^{k-2}\ell}+1)\cdots(2^{2^0\ell}+1)(2^\ell-1).\]Note that $2^\ell-1\equiv 1\pmod{3}$, so $3\nmid 2^\ell-1$. But $2^\ell-1\equiv -1\pmod{4}$ (note $\ell>1$), so it must have some prime factor $p\equiv -1\pmod{4}$, so $2^n-1$ has a prime factor $p\equiv -1\pmod{4}$, but $p\not=3$. This is a contradiction, so we must have $\ell=1$, as desired.

We claim that the only working $n$ are the powers of $2$. Let $p$ be a prime dividing $m^2 + 9$. By Fermat's Christmas Theorem, we must either have $p \mid 9$, in which case $p = 3$, or $p \equiv 1 \pmod{4}$. Hence, $2^n - 1$ is not divisible by any primes $p \equiv 3 \pmod{4}$ other than $3$. It is clear that $n = 1$ is a solution. Henceforth, suppose $n \geq 2$. Let $n = 2^k \cdot q$, where $q$ is odd. Then, we have \begin{align*} 2^n - 1 &= \left(2^q - 1\right)\left(2^q + 1\right)\left(2^{2q} + 1\right)\cdots\left(2^{(k - 1)q} - 1\right). \end{align*}Suppose $q > 1$. Since $q$ is odd, we must have $q \geq 3$. This implies that \begin{align*} 2^q - 1 &\equiv 3 \pmod{4}. \end{align*}Therefore, there must be a prime $p \equiv 3 \pmod{4}$ such that $p \mid 2^q - 1$. Additionally, since $q$ is odd, we have \begin{align*} 2^q - 1 &\equiv (-1)^q - 1 \pmod{3} \\ &\equiv 1 \pmod{3}. \end{align*}Hence $3 \nmid 2^q - 1$. Therefore, $p \neq 3$, contradiction. Hence, $q = 1$, so $n = 2^k$ is a power of $2$, as claimed. To finish, we prove that all of these solutions work. Write $n = 2^k$, where $k \in \mathbb{Z}_{\geq 0}$. For $k = 0, 1$, pick $m = 1, 3$ respectively. Henceforth, suppose $k > 1$. Then, \begin{align*} 2^{2^k} - 1 &= \prod_{i = 1}^{k - 1} \left(2^{2^i} + 1\right). \end{align*}We claim that the terms on the RHS are all relatively prime. Indeed, suppose $p \mid 2^{2^i} + 1$ for some $i$. We claim that $p \nmid 2^{2^j} + 1$ for all $j > i$. Noting that $p$ is odd, we have \begin{align*} 2^{2^i} &\equiv -1 \pmod{p} \\ \left(2^{2^i}\right)^{2^{j - i}} &\equiv (-1)^{2^{j - i}} \pmod{p} \\ 2^{2^j} &\equiv 1 \pmod{p} \\ &\not\equiv -1 \pmod{p}. \end{align*}Therefore, the terms on the RHS are all relatively prime, so by CRT it suffices to find $m$ satisfying \begin{align*} m^2 + 9 \equiv 0 \pmod{2^{2^i} + 1}, \end{align*}for all $i$, $1 \leq i \leq k - 1$. When $i = 1$, we have \begin{align*} m^2 + 9 &\equiv 0 \pmod{3} \\ m &\equiv 0 \pmod{3}. \end{align*}Write $m = 3t$. For $i > 1$, we have \begin{align*} m^2 + 9 &\equiv 0 \pmod{2^{2^i} + 1} \\ t^2 + 1 &\equiv 0 \pmod{2^{2^i} + 1} \end{align*}(division by $9$ makes sense here since $3 \nmid 2^{2^i} + 1$ for $i > 1$). This congruence is satisfied by picking \begin{align*} t &\equiv 2^{2^{i - 1}} \pmod{2^{2^i} + 1} \\ \implies m &\equiv 3 \cdot 2^{2^{i - 1}} \pmod{2^{2^i} + 1}. \end{align*}To summarize, we pick $m$ satisfying the following congruences: \begin{align*} m &\equiv 0 \pmod{3} \\ m &\equiv 3 \cdot 2^{2^{i - 1}} \pmod{2^{2^i} + 1}, \end{align*}where $2 \leq i \leq k - 1$. There exists such an $m$ by CRT. Thus, all $n$ of the form $2^k$ work, so we are done. $\Box$

1998 N5 wrote: Determine all positive integers $n$ for which there exists an integer $m$ such that ${2^{n}-1}$ is a divisor of ${m^{2}+9}$. Note that any prime $p$ dividing $m^2+9$ is either $3$ or $1 \pmod{4}$. Let $n=2^j\ell$ where $\ell$ is odd and $j \ge 0$. Then $2^{\ell}-1 \mid m^2+9$ and $3 \nmid 2^{\ell}-1$ hence all prime factors of $2^{\ell}-1$ are $1 \pmod{4}$ so $2^{\ell} \equiv 2 \pmod{4} \implies \ell=1$ and $n=2^j$. Now we show that any power of two works. Indeed, we have $2^{2^j}-1=3(2^{2}+1)(2^{4}+1) \dots (2^{2^{j-1}}+1)$ and all of the enlisted terms are pairwise coprime. Thus, if we can construct an $m$ modulo each of them, we are done. Suppose $m=3t$ and we want to pick $t$ such that $t^2+1 \equiv 0 \pmod{2^{2^s}+1}$ for all $s<j$. Since all prime factors of $2^{2^s}+1$ are $1 \pmod{4}$ this is indeed possible, and CRT concludes the proof. $\blacksquare$

The answer is when $n$ is a power of 2. The problem is just saying "find all $n$ such that $-9$ is a quadratic residue mod $2^n-1$". Claim 1: For an odd prime $p\neq 3$, $-9$ is a quadratic residue mod $p$ if and only if $p\equiv 1\pmod{4}$. If $p\equiv 1\pmod{4}$, then $\frac{p-1}{2}$ is even, so $$(-9)^{\frac{p-1}{2}}\equiv 3^{p-1}\equiv 1\pmod{p},$$so by Euler's criterion we have that $-9$ is a quadratic residue. On the other hand, if $\frac{p-1}{2}$ is odd, then it will evaluate to $-1$, so it is not a quadratic residue, hence proven. Claim 2: If $r$ is a quadratic residue mod $p$ for an odd prime $p$, then it is also a quadratic residue mod $p^m$ for all $m$. We will induct. Suppose that $$r\equiv k^2\pmod {p^m}.$$Then, note that $$(k+sp^m)^2\equiv k^2+2ksp^m+p^{2m}\equiv r+cp^m+p^m(2ks)=r+p^m(2ks+c)\pmod{p^{m+1}}$$for some unknown constant $c$. However, since $p$ is relatively prime to $2k$, we can choose $s$ so that $2ks+c\equiv 0\pmod{p}$, hence $r$ is a quadratic residue since some number of the form $(k+sp^m)^2$ works, shown by induction. Combining Claims 1 and 2 as well as the Chinese Remainder Theorem, $n$ works if and only if all prime divisors of $2^n-1$ are either 1 mod 4 or equal to 3. Now, if an odd prime $q$ divides $n$, then $2^q-1$, which is 3 mod 4, divides $2^n-1$, and it's also not divisible by 3, contradiction, so $n$ must be a power of 2. Now, we will just show that all powers of 2 work. Suppose FTSOC that $3\neq v\equiv3\pmod{4}$ is a prime that divides $2^{2^k}-1$. Then, by repeated difference of squares factorization, we have that $v\mid 2^{2^s}+1$ for some positive integer $s<k,$ since $v\neq 3.$ However, for any prime $v$ that divides $2^{2^s}+1$, the order of 2 mod $v$ has to be exactly $2^{s+1}$ (it has to be a power of 2, but if its a smaller power of 2 then $2^{2^s}$ is already 1 mod $v$). Hence, $v\equiv 1\pmod{2^{s+1}}$. However, since $s\geq 1$, this implies $v\equiv 1\pmod{4}$, contradiction, hence done.

Average FCT problem. The answer is all positive integers $n$ that are a power of $2$. Proof that $n$ must be a power of $2$: Let $k$ be the largest odd divisor of $n$. Assume for contradiction that $k>1$. By FCT, all prime divisors of $m^2+9$ are either $3$ or $1 \bmod{4}$, and since $2^k - 1$ is a divisor of $2^n - 1$, all prime divisors of $2^k-1$ are either $3$ or $1 \bmod{4}$. Since $3 \nmid 2^k-1$ and $2^k-1 \equiv 3 \pmod{4}$, it must have a prime divisor that is $3 \pmod{4}$, a contradiction. Thus $k=1$, and $n$ is a power of $2$. Powers of $2$ work: Let $n=2^j$. We prove the desired via induction on $j$, with the base cases $j=1, 2$ trivially true. For the inductive step, suppose $j$ works. Then factor \[ 2^{2^{j+1}}-1 = (2^{2^j}-1)(2^{2^j}+1). \]By inductive hypothesis the former factor works, and the latter works as it is $1 \bmod{4}$, so by the multiplicative nature of FCT we conclude.

We will show that the answer is all $n$ that are powers of $2.$ First suppose $p \mid n$ and $p \ne 2.$ Then there must be some prime $q \mid 2^p-1$ and $q \ne 3$ and $q \equiv 3 \pmod 4$ since $2^p-1 \equiv 3 \pmod 4.$ By Fermat christmas we have $q \mid m^2+9 \implies q \mid 9,$ contradiction. Now if $n=2^k$ we may write $2^{2^k}-1=(2+1)(2^2+1)(2^4+1)\dots(2^{2^{k-1}}+1).$ We notice that besides the first term, each term is a sum of two squares, so by Fermat christmas we have that all factors of this except for $3$ are $1 \pmod 4.$ Let one of these primes be $p.$ If we let $g$ be a primitive root $\pmod p,$ we may see that when $m\equiv 3g^{\frac{p-1}{4}} \pmod p$ we have $p \mid m^2+9,$ and when $m\equiv 0 \pmod 3$ we have $3 \mid m^2+9.$ Now taking all possible $p$ we get a system of congruences that must have a solution by CRT. Thus there exists such an $m,$ and we are done.

The answer is when $n$ is a power of $2$, including $n = 1$. Proof n must be a power of 2: Assume $n \geq 3$ is not a power of $2$. Then there exists an odd integer $k$ such that $$2^k-1 \mid 2^n - 1 \mid m^2 + 3^2.$$Since $2^k -1 \equiv 3 \pmod 4$, $2^k -1$ must have a $3 \pmod 4$ prime factor. From Fermat's Christmas Theorem, $2^k -1$ cannot have any $3 \bmod 4$ prime factors other than $3$. However, $$ 2^k-1 \equiv (-1)^k -1 \equiv 1 \pmod 3,$$so this is impossible. Therefore, $n$ must be a power of $2$. Proof n is a power of 2 works: Let $n = 2^k$, and $p$ be a prime such that $p \mid 2^{2^k} - 1$. Let the $a$ be the order of $2 \pmod p$. Then $a \mid \gcd(2^k, p-1).$ If $p \equiv 3 \pmod 4$, we must have $a = 2$, so $p = 3$. Therefore, $3$ is the only $3 \pmod 4$ prime factor of $2^{2^k} -1$, and all other prime factors are $1\pmod 4$. If $3^2 \mid 2^{2^k} - 1$, then we get $6 \mid 2^k$, since the order of $2 \pmod 9$ is $6$. This is clearly impossible, therefore $9 \nmid 2^{2^k}$. Thus, from Fermat's Christmas Theorem, there exists $m$ such that $2^{2^k} - 1 \mid m^2 + 9$.

nice $$\boxed{\text{The answer are all powers of 2.} }$$ Claim. $n$ must be a power of $2$ take a odd divisor $k>1$ of $n$, then $2^k-1\mid 2^n-1\mid m^2+9$ by Fermat Christmas Theorem all prime divisors of $m^2+9$ are $1\pmod 4$ or $3$, but $2^k-1\not\equiv 3\pmod 3$ as $k$ odd, nor $2^k-1\equiv 1 \pmod 4$, a contradiction. Claim. all powers of $2$ work. Let $n=2^k$, write $$2^{2^k}-1=(2+1)(2^2+1)(2^{2^2}+1)\cdots (2^{2^{k-1}}+1)$$Note that each term except for the first one, by Fermat Christmas Theorem have only $1\pmod 4$ prime divisors, so choose $g$ a primitive root $\pmod {p^\theta}$ then by Chinese Remainder Theorem exists $m$ such that $$m\equiv 3 g^{\frac{\phi({p^\theta})}{4}}\pmod {p^\theta} \hspace{0.4cm}\forall\hspace{0.2cm} p\mid 2^{2^j}+1 \hspace{0.5cm}\text{and} \hspace{0.5cm} m\equiv 0\pmod 3$$then $p^\theta\mid m^2+9$ , and $3\mid m^2+9$, as $\text{Fermat numbers}$ are relative primes we would have that $2^{2^n}-1\mid m^2+9$, as desired.

The answer is $\boxed{\text{powers of 2}}$. Let $d$ be the largest odd prime divisor of $n$. For the sake of contradiction, assume $d>1$. Fermat's Christmas Theorem states that the factors of $m^2+9$ are either $1 \pmod{4}$ or $3$. Since $2^d-1$ divides $2^n-1$, all the factors of $2^d-1$ are either $1 \pmod{4}$ or $3$. Then, as \[2^d-1 \equiv 3 \pmod{4},\] it must contain a factor that is $3 \pmod{4}$, contradicting our assumption that $d>1$. To prove the powers of $2$ work, induction with difference of squares easily works.

Our answer is $\boxed{n=2^b, b \ge 0}$. The most convenient method to show each works is by setting $m=3a$. Our condition is reduced to proving there exists such an $a$ with \[2^{2^b}-1 = 3\left(2^2+1\right)\left(2^{2^2}+1\right) \ldots \left(2^{2^{b-1}}-1\right) \mid 3(a^2+1) \mid (3a)^2+9.\] Since the Fermat primes are pairwise relatively prime, and we have the solution $a \equiv 2^{2^{t-1}} \pmod{2^{2^t}+1}$ for each $t$, there must exist such an $a$. Otherwise, suppose $p>1$ is an odd prime divisor of $n$. Then neither 2 or 3 are factors of $2^p-1 \equiv 3 \pmod 4$, so Fermat's Christmas Theorem tells us we must have $p \equiv 1 \pmod 4$ for all prime divisors of $n$, contradiction. $\blacksquare$

Since it's not specified that $m$ has to be an integer, the answer is obviously all positive integers $n$! (Just kidding! Real solution below) I claim that the answer is all $n$ that can be expressed in the form of $2^k$ for some nonnegative integer $k$. First, note that by Fermat's Christmas Theorem, if a prime $p$ divides $m^2+3^2$, then it is either $1$ mod $4$ or it divides $\gcd(m,3)$. Using this, I now claim that $n$ cannot have any prime factor $p$ that is larger than $2$. FTSOC, assume that $p\mid n$, where $p>2$. This then implies that \[2^p-1\mid 2^n-1,\]and since $p>2$, we have that $2^p-1$ must be $3$ mod $4$. Additionally, since $p$ is a prime $>2$, this means that $p$ is odd, implying that \[2^p-1 \equiv 2-1\equiv 1\mod 3,\]so if $2^p-1$ is $3$ mod $4$, some other prime $q\neq 3$ divides $2^p-1$, a contradiction to the Christmas Theorem statement. Therefore $n$ cannot have any prime factor $p>2$, meaning that $n$ must be in the form of $2^k$ in order for $2^n-1$ to have a multiple in the form of $m^2+9$. I now claim that for all $n=2^k$, $2^n-1$ has a multiple of the form $m^2+9$. I will prove this using induction. Suppose that $2^{2^k}-1$ for $k\geq 1$ has a multiple of the form $m^2+9$. This implies that $-9$ is a quadratic residue mod $2^{2^k}-1$. Now, note that \[2^{2^{k+1}}-1=(2^{2^k}-1)(2^{2^k}+1),\]and that $\gcd(2^{2^k}-1,2^{2^k}+1)=1$. Therefore, since we already know that $-9$ is a quadratic residue mod $2^{2^k}-1$, we just need to prove that $-9$ is also a quadratic residue mod $2^{2^k}-1$. Since the two modulos are relatively prime, by CRT, this will also prove that $-9$ is a quadratic residue mod $2^{2^{k+1}}-1$, meaning that a multiple of $2^{2^{k+1}}-1$ in the form of $m^2+9$. Using this, notice that, \[(2^{2^{k-1}})^2 \equiv -1 \mod (2^{2^{k}}+1) \iff (3*2^{2^{k-1}})^2 \equiv -9 \mod (2^{2^{k}}+1),\]which proves that $-9$ is indeed a quadratic residue mod $2^{2^{k}}+1$. Therefore, if there exists a multiple of $2^{2^{k}}+1$ in the form of $m_1^2+9$, then there exists a multiple of $2^{2^{k+1}}+1$ in the form of $m_2^2+9$! Finally, to complete our induction, we need to cover the base case of $n=2$ and the external case $n=1$. The latter is covered by $m=0$ and the former also by $m=0$. Therefore, there exists a multiple of $2^n-1$ in the form of $m^2+9$ if and only if $n$ is a power of $2$, finishing the problem.

We claim that the answer is $n=2^x$ for some nonnegative integer $x$. We start by proving that all other numbers fail. For any odd integer $n>1$, we have that $3\nmid 2^n-1$. We also have that $2^n-1\equiv 3\pmod{4}$, which implies that $2^n-1$ must have some prime divisor $p\equiv 3\pmod{4}$, and $p\neq3$. Since we require $2^n-1\mid m^2+9$, this prime must also divide $m^2+9$. However, we can then apply Fermat's Christmas theorem on $m^2+9$ to show that any prime $p$ dividing $m^2+3^2$ is either $p\equiv 1\pmod{4}$ or $p=3$, contradiction. Any even $n$ that is not a power of $2$ will have some odd factor $x>1$, and since $2^x-1\mid 2^n-1$ we get the same contradiction. Now we prove that $n=2^k$ works. $k=0$ gives $2^n-1=1$, which clearly works. We will now prove that for positive $k$, $2^{2^k}-1$ is the product of $3$ and some (not necessarily distinct) primes that are $1\pmod{4}$. We will use induction to do this, with base case $k=1$ and $2^{2^k}-1=3$. For the inductive step, we notice that $2^{2^{k+1}}-1=(2^{2^k}-1)(2^{2^k}+1)$, so we just need to prove that $2^{2^k}+1$ only has prime factors that are $1\pmod 4$. This is true by applying Fermat's Christmas Theorem, so the induction is complete. Now we just need to show there exists a working $m$. We can factor $2^n-1$ into powers of distinct primes, and consider each prime power separately. For the factor of $3$, just select $m\equiv0\pmod{3}$. For the other primes, which must have $p\equiv 1\pmod{4}$, we need to show that there exists a value of $m\pmod{p^k}$ such that $m^2+9\equiv\left(\frac m3 \right)^2+1\equiv 0 \pmod{p^k}$. It is well-known that there exists a primitive root $g$ with order $\phi(p^k)=(p-1)(p^{k-1})$. Since we know $\frac{p-1}{4}$ is an integer, we can let $\frac m3\equiv g^{\frac{(p-1)(p^{k-1})}{4}}\pmod{p^k}$ so that $\left(\frac m3 \right)^2\equiv -1 \pmod{p}$. Then we just multiply by $3$ to get a working value for $m\pmod{p^k}$. Then by applying CRT on all of the separate prime powers, we know that there will exist a value of $m$ such that $m^2+9$ is divisible by all of the prime powers together, and we are done.

The answer is all $n =2^k$ where $k$ is a nonnegative integer. We first introduce the following lemma. Lemma. For $n>1$, every prime divisor of $2^n-1$ is either $3$ or congruent to $1 \pmod 4$ if and only if $n$ is a power of $2$. Proof. First suppose that $n=2^k$, and let $p$ be a prime divisor of $2^n-1$. We have $$2^{2^k} \equiv 1 \pmod p$$so letting $\mathrm{ord}_p(2) = r$, it follows that $$\begin{cases} r \mid 2^k \\ r \mid p-1 \end{cases}$$so $r=2^l$ for $l \geq 1$. If $l=1$, then $4 \equiv 1 \pmod p$ or $p=3$; otherwise, $r \equiv 0 \pmod 4$, so we must have $4 \mid p-1$ or $p \equiv 1 \pmod 4$. Now, we show that if every prime divisor $p$ of $2^n-1$ is either $3$ or congruent to $1 \pmod 4$, then $n$ must be a power of $2$. For the sake of contradiction, suppose $n$ is not a power of $2$. Then, $n = 2^u \cdot w$ where $u$ is a nonnegative integer and $w$ is an odd integer greater than $1$. We know that $$2^w - 1 \mid 2^n - 1$$but $2^w - 1 \equiv (-1)^w - 1 \equiv 1 \not \equiv 0 \pmod 3$, so every prime dividing $2^w-1$ is congruent to $1 \pmod 4$. It is then clear that $$2^w - 1 \equiv 1 \pmod 4$$so $w=1$, a contradiction, giving the lemma. $\blacksquare$ Returning to the problem, suppose $q$ is a prime divisor of $2^n-1$. The given condition implies that $$m^2 \equiv -9 \pmod q$$so either $q=3$ or $$\left(\dfrac{-9}{q}\right) = \left(\dfrac{9}{q}\right) \left(\dfrac{-1}{q}\right) = \left(\dfrac{-1}{q}\right) = -1.$$It follows that $q \equiv 1 \pmod 4$, but $q$ is an arbitrary prime dividing $2^n-1$, so every prime dividing $2^n-1$ is congruent to $1 \pmod 4$. Applying the lemma finishes, since if $n$ is a power of $2$, we have $-9$ as a quadratic residue modulo every prime divisor of $n$ (or divisible by the prime divisor $3$), and the construction of $m$ follows from using the Chinese Remainder Theorem on the prime divisors of $n$. $\blacksquare$

Been a while since I wrote solutions. We claim that the answer is all $n$ of the form $\boxed {n=2^k}$. First note by Fermat's Two Squares Theorem that any prime factor of $m^2+9$ is either equal to $3$ or congruent to $1(mod \ 4)$. Now if $n$ is odd, it is trivial to observe that $2^n -1 \equiv 3(mod \ 4)$ for $n \geq 3$ so it follows that the required $n$ must be of the form $2^k$. We now show that these indeed work. Proceed by induction. We assume that $ 2^{2^k} -1 | m^2 +9$. Now we need to show that $2^{2^{k+1}} -1 | (m+a)^2 +9$ or that $a^2 +2am$ is divisible by $2^{2^k} -1$ and $2^{2^k} +1$. Since they are both coprime, the result follows by $CRT$ (This probably requires some care but good enough for a comeback I suppose).

great problem

The answer is $n=2^k$ where $k$ is a positive integer. We will use the following lemma: If $p\equiv 3\pmod 4$ and $p\mid x^2+y^2$, then $p\mid x$ and $p\mid y$. In particular, this means that $2^n-1$ has at most one prime factor that is $3$ modulo $4$, which is $3$. Suppose now that there is some prime number $p\mid n$, $p\equiv 3\pmod 4$. Clearly \[2^p-1\mid 2^n-1\mid m^2+9\]Notice now that $2^p-1\equiv 3\pmod 4$ so it must have a prime factor that is $3$ modulo $4$. This also can't be $3$ since $3\nmid 2^p-1$ (because $p$ is odd), hence we get a contradiction. Therefore, $n=2^k$ for some positive integer $k$. We prove that all such $n$ work. Note that we basically don't care about the $3$ in $2^{2^k}-1$ since from LTE $\nu_3\left(2^{2^k}-1\right)=1$ so we can just take $3\mid m$. Now I claim that the only prime factor that is $3$ modulo $4$ of $2^{2^k}-1$ is $3$. Proof: Note that \[2^{2^k}-1=(2+1)\left(2^2+1\right)\dots\left(2^{2^{k-1}}+1\right)\]Besides the first factor, all of them are of the form $x^2+1$ so they can't have a prime factor $p\equiv 3\pmod 4$. $\square$ For any other prime number $p\mid 2^{2^k}-1$ (note that from the above we have $p\equiv 1\pmod 4$) we have $\gcd(3,p)=1$ so $3\mid m$ doesn't affect us. Now note \[\left(\dfrac{-9}{q}\right)=\left(\dfrac{-1}{q}\right) \cdot \left(\dfrac{9}{q}\right)=1\]So there is some $m$ such that $p\mid m^2+9$. Therefore, by CRT we can find the desired $m$. $\blacksquare$

Does my proof of sufficiency work? The answer is $n = 2^k$. Assume $n > 1$ since $n = 1$ trivially holds. Necessity: Suppose that an odd number $d > 1$ divided $n,$ so that $2^d - 1$ divides $m^2 + 9$ and so $m^2 \equiv -9 \pmod{2^d - 1}.$ Since $d$ is odd, we have that $2^d - 1$ is not divisible by $3,$ and so we can further reduce this to $m^2 \equiv -1 \pmod{2^d - 1},$ i.e. $2^d - 1$ divides $m^2 + 1.$ We know by orders that $m^2 + 1$ only has prime divisors equivalent to $1 \pmod{4},$ so all divisors of $m^2 + 1$ are equivalent to $1 \pmod{4}.$ But $2^d - 1 \equiv 3 \pmod{4}$ for $d > 1,$ yielding a contradiction. Sufficiency: Let $n = 2^k.$ We will construct such a number $m$ inductively, with the base case $k = 1$ holding with $m = 0.$ Now assume that we have a number $m$ such that $$m^2 \equiv -9 \pmod{2^{2^k} -1}.$$Since $2^{2^{k+1}} - 1 = (2^{2^k} - 1)(2^{2^k} + 1)$ and $\gcd(2^{2^k} - 1, 2^{2^k} + 1) = 1,$ we just need to find a number $m$ such that $m^2 \equiv -9 \pmod{2^{2^k} + 1}$, from which we are done by the Chinese Remainder Theorem. Clearly $2^{2^k} + 1$ is not divisible by $3,$ so we can reduce this to $m^2 \equiv -1 \pmod{2^{2^k} + 1}.$ Factor $2^{2^k} + 1 = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}$ where the $p_i$ are primes and $e_i \in \mathbb{N}_0$ such that $p_i \equiv 1 \pmod{4}$ for all $i.$ By the Chinese Remainder Theorem, it suffices to show that $m^2 \equiv -1$ has a solution modulo $p_i^{e_i}$ for all $i.$ Notice that $p_i^{e_i}$ has a primitive root, and $\phi(p_i^{e_i}) = (p_i - 1)p_i^{e_i - 1}$ is divisible by $4,$ which readily implies that $-1$ is a quadratic residue. Combining all these congruences gives us our desired $m.$ Therefore, the only $n$ such that $2^n - 1$ has a multiple of the form $m^2 + 9$ are powers of two, which we have confirmed to work.

Suppose that $k\mid n$ for some odd $k>2$. If $m^2\equiv-9\pmod{2^n-1}$ for some $m$, then $m^2\equiv-9\pmod{2^k-1}$. Since $2^k\equiv2\pmod3$, we have $(\frac{m}3)^2\equiv-1\pmod{2^k-1}$. Since $2^k-1\equiv3\pmod4$, it follows that $-1$ is a quadratic residue modulo $p$ for some prime divisor $p$ such that $p\equiv3\pmod4$, a contradiction. Hence, if $n$ is not a power of $2$, then $2^n-1$ has no multiple of the form $m^2+9$. Suppose that $-9$ is a quadratic residue modulo $2^{2^r}-1$ for some nonnegative integer $r$. note that $(3\cdot 2^{2^{r-1}})^2\equiv-9\pmod{2^{2^r}+1}$. Then $-9$ is a quadratic residue modulo $(2^{2^r}-1)(2^{2^r}+1)=2^{2^{r+1}}-1$. By induction, it follows that $-9$ is a quadratic residue modulo $2^{2^r}-1$ for all $r$, so that $2^{2^r}-1$ has a multiple of the form $m^2+9$. We thus conclude that the solution set for $n$ is given by $\boxed{\{2^r\mid r\in\mathbb{Z},r\ge0\}}.$ $\blacksquare$