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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Geometry
gggzul   0
19 minutes ago
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
0 replies
gggzul
19 minutes ago
0 replies
hard problem
Cobedangiu   5
N 27 minutes ago by KhuongTrang
$a,b,c>0$ and $a+b+c=7$. CM:
$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+abc \ge ab+bc+ca-2$
5 replies
Cobedangiu
Yesterday at 4:24 PM
KhuongTrang
27 minutes ago
Nordic 2025 P3
anirbanbz   9
N 30 minutes ago by Tsikaloudakis
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
9 replies
anirbanbz
Mar 25, 2025
Tsikaloudakis
30 minutes ago
Aime type Geo
ehuseyinyigit   1
N an hour ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
1 reply
ehuseyinyigit
Yesterday at 9:04 PM
ehuseyinyigit
an hour ago
No more topics!
IMO ShortList 1998, number theory problem 5
orl   64
N Apr 25, 2025 by Ilikeminecraft
Source: IMO ShortList 1998, number theory problem 5
Determine all positive integers $n$ for which there exists an integer $m$ such that ${2^{n}-1}$ is a divisor of ${m^{2}+9}$.
64 replies
orl
Oct 22, 2004
Ilikeminecraft
Apr 25, 2025
IMO ShortList 1998, number theory problem 5
G H J
Source: IMO ShortList 1998, number theory problem 5
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AlanLG
241 posts
#56 • 1 Y
Y by cubres
nice :blush: $$\boxed{\text{The answer are all powers of 2.} }$$ 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.
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joshualiu315
2534 posts
#57 • 1 Y
Y by cubres
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.
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shendrew7
795 posts
#58 • 1 Y
Y by cubres
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. $\blacksquare$
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peppapig_
281 posts
#60 • 1 Y
Y by cubres
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.
This post has been edited 2 times. Last edited by peppapig_, Mar 14, 2024, 12:48 AM
Reason: Parentheses
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SenorSloth
37 posts
#61 • 2 Y
Y by OronSH, cubres
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.
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blueberryfaygo_55
340 posts
#62 • 2 Y
Y by megarnie, cubres
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 $n$. $\blacksquare$
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Eka01
204 posts
#63 • 1 Y
Y by cubres
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).
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onyqz
195 posts
#64 • 1 Y
Y by cubres
great problem
solution
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SomeonesPenguin
128 posts
#65 • 1 Y
Y by cubres
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 $m$. $\blacksquare$
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EpicBird08
1751 posts
#67 • 1 Y
Y by cubres
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.
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smileapple
1010 posts
#68 • 1 Y
Y by cubres
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 $\boxed{\{2^r\mid r\in\mathbb{Z},r\ge0\}}.$ $\blacksquare$
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cursed_tangent1434
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#69 • 1 Y
Y by cubres
Mid problem, but I think it's quite instructive. Doesn't help if you misread and spend a non-trivial amount of time trying to solve the problem with 'divisor' instead of 'multiple'.

We claim that the answer is all positive integers $n$ of the form $n=2^k$ for some non-negative integer $k$. When $n=1$ the result is clear so we deal with $n>1$ in what follows. First we shall show that all the claimed solutions indeed work.

Clearly $3 \mid 2^{2^k}-1$ and in particular, Lifting the Exponent Lemma tells us that $\nu_3(2^{2^{k}}-1)=\nu_3(3)+\nu_3(2^{k-1})=1$. Thus, we can write
\[2^n-1=3 \cdot p_1^{a_1}\cdot p_2^{a_2}\cdots p_r^{a^r}\]for distinct odd primes $p_1, p_2, \cdots , p_r$. We first claim the following.

Claim : All primes $p_1,p_2,\dots , p_r$ are $1 \pmod{4}$.

Proof : Simply note that since
\[2^{2^k}-1=(2-1)(2+1)(2^2+1)\dots (2^{2^{k-1}}+1)\]Since each index of $2$ beyond the second factor is even, none of these factors are divisible by 3. Further, if $p \equiv 3 \pmod{4}$ (and $p>3$) divides one of these terms, then $p \mid 2^{2^i}+1$ for some $i \ge 1$. However, this is a clear contradiction by Fermat's Christmas Theorem implying that the left-hand expression has no $3 \pmod{4}$ primes factors as claimed.


We let $3 \mid m$. Further, for each prime $p_i \mid 2^{2^k}-1$ we have,
\[(-9)^{\frac{p_i-1}{2}} \equiv (-1)^{\frac{p_i-1}{2}}\cdot 9^{\frac{p_i-1}{2}} \equiv (-1)^{\frac{p_i-1}{2}} \equiv 1 \pmod{p_i}\]since $4 \mid p_i-1$ as noted above. Thus, $-9$ is a QR $\pmod{p_i}$. Thus, for each prime $p_i$ there exists some positive integer $x_i$ for which $p_i \mid x_i^2+9$. We now resort to induction.

Say there exists a positive integer $x_{im}$ such that $p^m \mid x_{im}^2+9$ for some $m \ge 1$. We note that,
\[(x_{im}+kp^m)^2+9\]is injective $\pmod{p^{m+1}}$ and in the set $\{0,p^m,2\cdot p^m,\dots (p-1)\cdot p^m\}$ as $k$ ranges from $0$ to $p-1$. Thus, one of these values for $k$ gives a new positive integer $x_{i(m+1)}=x_{im}+kp^m$ such that $p^{m+1}\mid x_{i(m+1)}^2+9$.

Now, applying the inductive described above to $p_i$, we conclude that there must exist some positive integer $y_i$ such that $p_i^{a_i} \mid y_i^2 +9$ for all $1 \le i \le r$. We apply CRT on $3 , p_1,p_2,\dots , p_r$ to construct a suitable value for $m$ and finish.

To see why no other $n$ work, say $p \mid n$ and $2<p<n$. Thus,
\[2^p-1\mid 2^n-1 \mid m^2+9\]Since $p>2$ the left-hand side is $3 \pmod{4}$ and thus, some prime factor of $2^p-1$ , $q \ne 9 \equiv 3 \pmod{4}$ exists. However, by Fermat's Christmas Theorem this is a clear contradiction as we must have $q \mid 9$.
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AshAuktober
1004 posts
#70 • 1 Y
Y by cubres
Odd $n$ don't work so $n$ must be a power of 2. Such $n$ can be shown to work using CRT combined with LTE to show $\nu_3$ can't be too large.

Also why is misreading this so real lmfao.
This post has been edited 1 time. Last edited by AshAuktober, Apr 10, 2025, 12:18 PM
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ATM_
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#71 • 1 Y
Y by cubres
Suppose $n$ has an odd prime divisor d , such that : $d>1$
We have : $2^d-1|2^n-1$
Hence : $2^d-1|a^2+9$
$d>1\implies 2^d-1>1$ , thus: $2^d-1$ has an odd prime divisor p
By Fermat Christmas theorem: $p \equiv 1[4]$
Hence : $p|a^2+9\implies a^2\iff -9[p]$
Using quadratic recopricity : $\left(\dfrac{-9}{p}\right)=\left(\dfrac{-1}{p}\right).\left(\dfrac{3}{p}\right)^2=-1.\left(\dfrac{3}{p}\right)^2$
Because : $p\iff 3[4]$

If: $p\neq 3$ , then:$ \left(\dfrac{3}{p}\right)=\pm 1\implies \left(\dfrac{3}{p}\right)^2=1$
So :b$ \left(\dfrac{-9}{p}\right)=-1$ (not a quadratic residu mod p)
Absurde , Hence: $p=3$

So: $3|2^d-1\implies 2^d\equiv 1[3]$
Hence : $o_p(2)|d\implies 2|d$
Absurde ,cause d is odd
Thus n has no odd divisors
Which means $n$ is a power of 2
Let : $n=2^k/k\in \mathbb{N}$

Proof by induction:
For $n=1,2^n-1=1|a^2+9$
True
Suppose :$\exists a\in \mathbb{N}:a^2\equiv -9[2^{2^{k}}-1]$

We habe : $2^{2^{k+1}}-1=(2^k-1)(2^k+1)$
For some positive integer $a_1=3×2^{2^{k-1}},with :k\in \mathbb{N}*$
We have : $a_1^2+9=9×2^{2^k}+9=9(2^{2^k}+1)$
Thus : $2^{2^k}+1|a_1^2+9$
And : $a^2\equiv -9[2^{2^k}-1]$
Because : $pgcd(2^{2^k}-1,2^{2^k}+1)=gcd(2^{2^k}+1,2)=1$
By CRT: $\exists b\in \mathbb{N}:b\equiv a_1[2^{2^k}+1]$ and $b\equiv a[2^{2^k}-1]$
$\implies b^2\equiv a^2\equiv -9[2^{2^k}-1]$ et $b^2\equiv a_1^2\equiv -9[2^{2^k}+1]$
So : $\exists b\in \mathbb{N}:b^2\equiv -9[2^{2^{k+1}}-1]$

Hence : $\exists a\in \mathbb{N}:2^n-1|a^2+9$ iff $n$ is a power of 2
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Ilikeminecraft
616 posts
#72
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We are solving when $2^n - 1\mid m^2 + 9.$ I claim that an integer $d$ has a multiple of the form $m^2 + 9$ if and only if none of $d$'s prime factors, other than $3$, are $3\pmod4$, and $\nu_3(d) \leq 2.$

We first show that all $d$ that has a multiple of the form $m^2 + 9$ must be of that form. By Fermat's Christmas theorem, we have that all of the prime divisors of $m^2 + 9$ must be $1\pmod4,$ or is 2. Now, assume that $\nu_3(d) \geq 3.$ Thus, any multiple of $d$ divides 27. However, $\nu_3(m^2 + 9) \leq2.$ Thus, this direction is now proved.

Now, we show that all $d$ of this form has a multiple that can be written as $m^2 + 9.$ By using the Jacobi Symbol, we can use quadratic reprocity to show that there exists an $m$ such that $m^2 \equiv -1\pmod d,$ and thus we are done.

Now, we are looking for when $2^{n} - 1$ satisfies the conditions of $d.$ For all the prime factors to be $1\pmod4$ or $3$ we claim that $n$ is a power of 2. This can be seen by using orders.

Since $\nu_3(2^{k - 1}) = 0,$ by $LTE,$ our expression has at most one power of 3. Thus, the answer is $n = \boxed{2^k}$ for $k\in\mathbb N$
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