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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Inspired by 2025 Beijing
sqing   11
N 2 minutes ago by sqing
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
11 replies
1 viewing
sqing
Saturday at 4:56 PM
sqing
2 minutes ago
A functional equation
super1978   1
N 19 minutes ago by CheerfulZebra68
Source: Somewhere
Find all functions $f: \mathbb R \to \mathbb R$ such that:$$ f(f(x)(y+f(y)))=xf(y)+f(xy) $$for all $x,y \in \mathbb R$
1 reply
super1978
26 minutes ago
CheerfulZebra68
19 minutes ago
Prove that IMO is isosceles
YLG_123   4
N 2 hours ago by Blackbeam999
Source: 2024 Brazil Ibero TST P2
Let \( ABC \) be an acute-angled scalene triangle with circumcenter \( O \). Denote by \( M \), \( N \), and \( P \) the midpoints of sides \( BC \), \( CA \), and \( AB \), respectively. Let \( \omega \) be the circle passing through \( A \) and tangent to \( OM \) at \( O \). The circle \( \omega \) intersects \( AB \) and \( AC \) at points \( E \) and \( F \), respectively (where \( E \) and \( F \) are distinct from \( A \)). Let \( I \) be the midpoint of segment \( EF \), and let \( K \) be the intersection of lines \( EF \) and \( NP \). Prove that \( AO = 2IK \) and that triangle \( IMO \) is isosceles.
4 replies
YLG_123
Oct 12, 2024
Blackbeam999
2 hours ago
Geometric mean of squares a knight's move away
Pompombojam   0
2 hours ago
Source: Problem Solving Tactics Methods of Proof Q27
Each square of an $8 \times 8$ chessboard has a real number written in it in such a way that each number is equal to the geometric mean of all the numbers a knight's move away from it.

Is it true that all of the numbers must be equal?
0 replies
Pompombojam
2 hours ago
0 replies
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
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Source: IMO ShortList 1998, number theory problem 5
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orl
3647 posts
#1 • 11 Y
Y by Davi-8191, pog, HWenslawski, rg_ryse, Adventure10, jhu08, megarnie, mathmax12, Mango247, WiseTigerJ1, cubres
Determine all positive integers $n$ for which there exists an integer $m$ such that ${2^{n}-1}$ is a divisor of ${m^{2}+9}$.
Attachments:
This post has been edited 1 time. Last edited by orl, Oct 23, 2004, 12:41 PM
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orl
3647 posts
#2 • 7 Y
Y by pog, Adventure10, rg_ryse, jhu08, Mango247, WiseTigerJ1, cubres
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 :)
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grobber
7849 posts
#3 • 19 Y
Y by numbertheorist17, SupermemberVN, randomasdf97, baopbc, hmida99, Polynom_Efendi, Williamgolly, pog, Adventure10, rg_ryse, jhu08, Mango247, MarioLuigi8972, WiseTigerJ1, Stuffybear, ehuseyinyigit, cubres, and 2 other users
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. $2^n-1|m^2+9$.
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abdurashidjon
119 posts
#4 • 6 Y
Y by pog, Adventure10, jhu08, WiseTigerJ1, cubres, and 1 other user
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
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ZetaX
7579 posts
#5 • 6 Y
Y by pog, Adventure10, jhu08, WiseTigerJ1, cubres, and 1 other user
And what do you want to tell us by this¿
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Davron
484 posts
#6 • 6 Y
Y by pog, Adventure10, jhu08, Mango247, WiseTigerJ1, cubres
i think he wants the proofs for this facts

davron
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Yimin Ge
253 posts
#7 • 8 Y
Y by pog, Adventure10, jhu08, looosp, Hexagrammum16, Mango247, WiseTigerJ1, cubres
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:

Click to reveal hidden text
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KingSmasher3
1399 posts
#8 • 6 Y
Y by pog, jhu08, TomMae10, Adventure10, WiseTigerJ1, cubres
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, $a \equiv b \equiv 0\pmod{4m+3}.$
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iarnab_kundu
866 posts
#9 • 6 Y
Y by pog, jhu08, Adventure10, Mango247, WiseTigerJ1, cubres
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 $t$.
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KingSmasher3
1399 posts
#10 • 5 Y
Y by pog, jhu08, Adventure10, WiseTigerJ1, cubres
Alternative Solution with Legendre's Symbol
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Eray
381 posts
#11 • 7 Y
Y by pog, jhu08, joseph02, Adventure10, Mango247, WiseTigerJ1, cubres
$2^n - 1 \mid m^2 + 9$

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.
This post has been edited 1 time. Last edited by Eray, Oct 10, 2014, 8:44 AM
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TheOverlord
97 posts
#12 • 6 Y
Y by pog, jhu08, Adventure10, Mango247, WiseTigerJ1, cubres
Eray wrote:
For $k\ge3$, $2^{2^3}-1\mid 2^{2^k}-1$. So $127\mid 2^{2^k}-1$
But $2^{2^3}-1=255$!!!
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Eray
381 posts
#13 • 5 Y
Y by pog, jhu08, Adventure10, WiseTigerJ1, cubres
I have corrcted my solution. Thank you @TheOverlord
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sayantanchakraborty
505 posts
#14 • 5 Y
Y by pog, jhu08, Adventure10, WiseTigerJ1, cubres
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 $n=1,2^k \forall k \in \mathbb{Z^+}$
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cobbler
2180 posts
#15 • 5 Y
Y by pog, jhu08, Adventure10, WiseTigerJ1, cubres
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 $n=2^a$.
This post has been edited 1 time. Last edited by cobbler, Mar 25, 2015, 10:50 PM
Reason: A silly
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