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jwelsh   0
Jul 1, 2025
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1 viewing
jwelsh
Jul 1, 2025
0 replies
A geometry problem
Lttgeometry   0
27 minutes ago
Let \( \triangle ABC \) be an acute triangle, and let \( I \) be its incenter. Let \( \omega \) be the incircle of triangle \( ABC \), which touches \( BC, CA, AB \) at points \( D, E, F \), respectively. Let \( AD \) intersect \( \omega \) again at point \( J \ne D \), and let the circumcircle of triangle \( BCJ \) intersect \( \omega \) again at point \( K \ne J \). The circumcircles of triangles \( BFK \) and \( CEK \) intersect again at a point \( L \ne K \). Let \( M \) be the midpoint of arc \( BAC \). Prove that points \( M, I, L \) are collinear.
0 replies
Lttgeometry
27 minutes ago
0 replies
Peru Ibero TST 2022
diegoca1   0
38 minutes ago
Source: Peru Ibero TST 2022 D2 P2
On a $2 \times n$ board, we call a bar each segment on the border of the board that is adjacent to some cell. The $2n + 4$ bars are numbered from $1$ to $2n + 4$ (as shown in the figure).
(See figure 1)
A diagonal is drawn in each cell. These diagonals divide the board into several pieces. For each bar, there is a unique other bar such that both belong to the boundary of the same piece. In this way, a pairing of the $2n + 4$ bars is formed. For example, in the figure below, the pairing is $(1, 10), (2, 5), (3, 4), (6, 7), (8, 9)$.
(See figure 2)
On a $2 \times 9$ board, how many possible pairings are there?
Note: Two pairings with the same pairs (regardless of their order) are considered the same. For example, the following pairings are considered identical:
\[
(1,10), (2,5), (3,4), (6,7), (8,9)
\quad \text{and} \quad
(2,5), (7,6), (9,8), (10,1), (3,4).
\]
0 replies
diegoca1
38 minutes ago
0 replies
Students in a Bookstore
TigerOnion   0
41 minutes ago
Source: Cyprus IMO First TST 2014
On the last Friday before the holidays, at a certain school, many students decided to visit the school library. Each student arrived at the library and left only once. However, it was observed that for any three students, at least two of them were present in the library at the same time (i.e., their time intervals in the library overlapped). Prove that there exist two time points during the day such that every student was present in the library for at least one of them.
0 replies
TigerOnion
41 minutes ago
0 replies
Peru Ibero TST 2022
diegoca1   0
an hour ago
Source: Peru Ibero TST 2022 D2 P1
For every positive integer $m > 1$, let $p(m)$ be the largest prime number that divides $m$. For $m = 1$, define $p(1) = 1$.

a) Prove that there exists a positive integer $n$ such that the numbers $p(n - 2022)$, $p(n)$, and $p(n + 2022)$ are all less than $\frac{\sqrt{n}}{20}$.

b) Given a positive integer $N$, prove that there exists a positive integer $n$ such that the numbers $p(n - 2022)$, $p(n)$, and $p(n + 2022)$ are all less than $\frac{\sqrt{n}}{N}$.
0 replies
diegoca1
an hour ago
0 replies
No more topics!
Same divisor
sam-n   16
N May 22, 2025 by AbdulWaheed
Source: IMO Shortlist 1997, Q14, China TST 2005
Let $ b, m, n$ be positive integers such that $ b > 1$ and $ m \neq n.$ Prove that if $ b^m - 1$ and $ b^n - 1$ have the same prime divisors, then $ b + 1$ is a power of 2.
16 replies
sam-n
Mar 6, 2004
AbdulWaheed
May 22, 2025
Source: IMO Shortlist 1997, Q14, China TST 2005
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sam-n
793 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ b, m, n$ be positive integers such that $ b > 1$ and $ m \neq n.$ Prove that if $ b^m - 1$ and $ b^n - 1$ have the same prime divisors, then $ b + 1$ is a power of 2.
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grobber
7849 posts
#2 • 4 Y
Y by Mathotsav, Adventure10, Mango247, and 1 other user
b<sup>n</sup>-1 is a Mersenne sequence, meaning that (b<sup>m</sup>-1, b<sup>n</sup>-1)=b<sup>(m,n)</sup>-1. Let d=(m,n). This means that b<sup>d</sup>-1 and b<sup>m</sup>-1 have the same prime divisors. Let b<sup>d</sup>=a, and m/d=M. Then a-1 and a<sup>M</sup>-1 have the same prime divisors.

Let p|a<sup>M</sup>-1 be a prime. Then p|a-1, so p|a<sup>M</sup>-a=a(a<sup>M-1</sup>-1) so p|a<sup>M-1</sup>. We continue like this until we get p|a<sup>2</sup>-1=(a-1)(a+1). We get that a-1 and (a-1)(a+1) have the same prime divisors. It's easy to get from here a+1=2<sup>t</sup>, so b<sup>d</sup>+1=2<sup>t</sup>. If d is even then b<sup>d</sup> is a perfect square which is 3(mod 4), which is false, so d is odd, so 2<sup>t</sup>=b<sup>d</sup>+1 is divisible with b+1, so b+1 is a power of 2.
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grobber
7849 posts
#3 • 6 Y
Y by mutasimmimxx, math4444, Adventure10, Mango247, MarioLuigi8972, and 1 other user
I'll use a lemma: if $a>1,k>1$ and $a^k-1,a-1$ have the same prime factors, then $k$ is a power of $2$ and so is $a+1$.

Assume the contrary, and let $p$ be an odd prime factor of $k$. Then any prime factor $q$ of $1+a+\ldots+a^{p-1}$ also divides $1+a+\ldots+a^{k-1}|a^k-1$, so it divides $a-1$, meaning that $1+a+\ldots+a^{p-1}\equiv p\pmod q\equiv 0\pmod q$, so $q=p$, so $1+a+\ldots+a^{p-1}=p^t,\ t>1$. Put $a=up+1$ and we'll get a contradiction fast (we find $1+a+\ldots+a^{p-1}\equiv p\pmod{p^2}$). This shows that $k$ is a power of $2$. If we take $r$ to be a prime which divides $a+1$, then it will divide $a^k-1$, and thus $a-1$, so $r=2$, which is what we wanted.

Apply the lemma for $a=b^d$, where $d=(m,n)$, and $k=\frac md$. We find that all prime factors of $b^{kd}-1$ also divide $b^d-1$, where $k$ is a power of $2$. We also know that $b^d+1=2^t,\ t\ge 2$. Since a perfect square cannot be equal to $2^t-1,\ t\ge 2$, it means that $d$ is odd. Take a prime $r|b+1|b^2-1|b^{kd}-1$, so $r|b^d-1\equiv -2\pmod r$, so $r=2$, meaning that $b+1$ is a power of $2$, Q.E.D.

P.S. The fact that $k$ is a power of $2$ is only useful for showing that $kd$ is even.
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Pascual2005
1160 posts
#4 • 2 Y
Y by Adventure10, Mango247
It is nice and easy: One can show that if the prime factors of $A^n-1$ and $A^m-1$ coincide, then $n=2, m=1$ and $A+1=2^t$. And also using this one can show that If the prime factors of $A^m-1$ are a subset of the ones from $A^n-1$ then $m|n$ or $m=2$ and $A+1=2^t$ and a list of nice corollaries follows:

The nicest
Corollary:

$A_1^{x_1}-A_2^{x_2} A_2^{x_3}....A_n^{x_n}=1$ with $A_i$ given integers has at most one solution $(x_1,x_2...,x_n)$ with $x_1>1$ and $x_i>0$.

For example $3^x-2^y7^z=1$ has at most one solution $(x,y,z)$
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Robert_89
3 posts
#5 • 2 Y
Y by Adventure10, Mango247
Lets say m>n

Because
\[ (b^m - 1 , b^n - 1 ) = b^{(m,n)} - 1
\]
then $ b^{(m,n)} - 1$ and $ b^m - 1$ must have same prime divisors.
so now put $ (m,n) = \zeta , m = k \zeta , a = b^\zeta$ so now it is clear that $ a - 1 , a^k - 1$ share same prime divisors.
now suppose $ P$ is one of $ k$ prime divisors.
because $ a^p - 1 \mid a^k - 1 , a - 1 \mid a^p - 1$ , $ a - 1, a^p - 1$ also must have same prime divisors.
Now lets put : $ \alpha = a^{p - 1} + a^{p - 2} + ... + 1$
If $ q$ be a prime divisor of $ \alpha$ , Then $ a - 1 \equiv 0 (mod q)$ , as a result, $ \alpha \equiv 1 + 1 + 1 + ... + 1 \equiv p$ so that means $ q = p$
and the obviously we can say that $ \alpha = p ^ \phi$
in order to find out what $ \phi$ can be , we write : $ a = pt + 1$ ( $ a - 1 \equiv 0, (mod p )$ )
now if we look closer we easily see that :
$ a^r \equiv (pt + 1)^r \equiv \sum(j = 0 ,\rightarrow r) (pt)^j * C(r,j) \equiv C(r,1)pt + 1 (mod p^2 )$
$ \Rightarrow \alpha \equiv \sum (r = 0 \rightarrow p - 1) rpt + 1 \equiv pt \frac {(p - 1)p} {2} + p \equiv p (mod p^2)$

now if $ p$ is prime and odd, $ \phi = 1 \rightarrow = a^p - 1 + ... + 1 = p$ and this contradicts $ a > 1$ , therefore p should be even, and that is $ p = 2$
so prime divisors of $ a^2 - 1 , a - 1$ are the same. but $ (a + 1) - (a - 1) = 2$
and so $ a + 1 = 2^ w$ ,,, inother words , $ b^\zeta + 1$ is a power of $ 2$ ,
and again ofcourse $ \zeta$ cant be even , because if be, then $ b^\zeta + 1 \equiv 2 (mod 4)$ and because $ b^\zeta + 1 = 2 ^ w$ we would conclude that $ b = 1$ (since b is odd)
therefore $ \zeta$ is odd and $ b + 1 \mid b^\zeta + 1$
and we are all set, $ b + 1$ is a power of $ 2$ .
:ninja:
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mai quoc thang
66 posts
#6 • 1 Y
Y by Adventure10
A good solution ,Robert_89 , but there are too many mistakes in your solution

I had tried my best to understand your solution

In order to take profits from Mathlinks . You should pratise English more . :lol:
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Zhero
2043 posts
#7 • 7 Y
Y by Binomial-theorem, K.N, a22886, Adventure10, Mango247, panche, and 1 other user
Without loss of generality, suppose $m < n$. If $b+1$ is not a power of 2 and $n \neq 6$, then by Zsigmondy's theorem we can find some prime $p$ such that $p | b^n - 1$ but $p \not | b^m - 1$, which is impossible. Furthermore, $2^1 - 1 = 1$, $2^2 - 1 = 3$, $2^3 - 1 = 7$, $2^4 - 1 = 3 \cdot 5$, $2^5 - 1 = 31$, and $2^6 - 1 = 3^2 \cdot 7$, so $2^6 - 1$ does not have the same set of prime factors as any of $2^i - 1$ with $1 \leq i \leq 5$. Hence, $n \neq 6$, so we may conclude that $b+1$ is a power of 2.

Note that Zsigmondy's theorem tells us that $b+1$ must be a power of two and that $(m,n) = (1,2)$ or $(m,n) = (2,1)$.
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Aimingformygoal
126 posts
#8
Y by
sam-n wrote:
Let $ b, m, n$ be positive integers such that $ b > 1$ and $ m \neq n.$ Prove that if $ b^m - 1$ and $ b^n - 1$ have the same prime divisors, then $ b + 1$ is a power of 2.

Solution
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hydo2332
435 posts
#9
Y by
Let $d = (m,n)$, with $m = dm_0, n = dn_0$ and $(m_0,n_0) = 1$. We have that $(b^n - 1, b^m - 1) = b^d - 1$, so we get that

$b^{dm_0} - 1 = (b^d - 1)\cdotM$
$b^{dn_0} - 1 = (b^d - 1)\cdotN$
Notice that if there is a new prime $p$ that divides $M$ but does not divide $b^d -1$, then it divides $b^{dm_0} - 1$, which due to the problem implies that this prime divides $b^{dn_0} - 1$. But this is would implie that $p$ divides $(b^{dm_0} - 1, b^_{dn_0} - 1) = b^d - 1$, which is a contradiction.
Hence, every prime that divides $b^d - 1$ must divide $b^{dm_0} - 1$ and vice-versa.
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Aliosman
11 posts
#10
Y by
Bruhhh direct zsigmondy;
Let $m>n$ from zsigmondy's theorem there is must be non-common prime divisor. But zsigmondy does not work at these two case
Case 1: $b=2$ and $n=6$ trying for $m=1,2,3,4,5$ we get that no sol.
Case 2: $b+1=2^i$ from this statement we're done.
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hectorleo123
368 posts
#11
Y by
sam-n wrote:
Let $ b, m, n$ be positive integers such that $ b > 1$ and $ m \neq n.$ Prove that if $ b^m - 1$ and $ b^n - 1$ have the same prime divisors, then $ b + 1$ is a power of 2.
WLOG $m>n$:
$gcd(b,1)=1$
By Zsigmondy's Theorem:
$\exists$ prime $p/ p|b^m-1, p\nmid b^n-1$
$(\Rightarrow \Leftarrow)$
But there is an exception:
$b+1=2^t_\blacksquare$
This post has been edited 1 time. Last edited by hectorleo123, May 18, 2023, 12:38 AM
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smileapple
1015 posts
#12
Y by
easy one ;)

Without losing generality, suppose that $m<n.$ Since $b^m-1$ and $b^n-1$ share the same prime divisors, there does not exist some prime $p$ such that $p\mid b^n-1$ and $p\nmid b^i-1$ for all $i\in\{1,2,\cdots,n-1\}$, as $m<n$ implies that $m\in\{1,2,\cdots,n-1\}.$ Thus, the triplet $(b,1,n)$ must be an exception to Zsigmondy's Theorem, implying that either $b+1$ is a power of $2$ and $n=2$ or $(b,n)=(2,6).$

If we have the latter case, then $b^n-1=2^6-1=63=3^2\cdot7.$ One can check that there does not exist an element of $\{2^1-1,2^2-1,\cdots,2^5-1\}$ such that its prime divisors are exactly $3$ and $7$. Thus this case fails, so it follows that $b+1=2^k$ for some positive integer $k$ and $n=2$, as desired. $\blacksquare$
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MELSSATIMOV40
30 posts
#13 • 1 Y
Y by shafikbara48593762
Zsygmondy's theorem kills it
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Eka01
204 posts
#14
Y by
Just use zsigmondy
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Frank25
1 post
#15
Y by
Direct application of Zsigmondy theorem.
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SlovEcience
30 posts
#16
Y by
How do you prove Zsigmondy's theorem?
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AbdulWaheed
34 posts
#17
Y by
Knowing Zsigmondy's theorem, is enough to deal b+1.Might this fancy theorem again appears in IMO 2025.
Wishing
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