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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Orthocenter
jayme   4
N 19 minutes ago by Sadigly
Dear Mathlinkers,

1. ABC an acuatangle triangle
2. H the orthcenter of ABC
3. DEF the orthic triangle of ABC
4. A* the midpoint of AH
5. X the point of intersection of AH and EF.

Prove : X is the orthocenter of A*BC.

Sincerely
Jean-Louis
4 replies
jayme
Mar 25, 2015
Sadigly
19 minutes ago
Concurrency
Dadgarnia   29
N 21 minutes ago by blueprimes
Source: Iranian TST 2020, second exam day 2, problem 4
Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent.

Proposed by Alireza Dadgarnia
29 replies
Dadgarnia
Mar 12, 2020
blueprimes
21 minutes ago
Good Numbers
ilovemath04   30
N 35 minutes ago by ihategeo_1969
Source: ISL 2019 N5
Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\tbinom{an}{b}-1$ is divisible by $an+1$ for all positive integers $n$ with $an \geq b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime.
30 replies
ilovemath04
Sep 22, 2020
ihategeo_1969
35 minutes ago
Sequences problem
BBNoDollar   1
N an hour ago by BBNoDollar
Source: Mathematical Gazette Contest
Determine the general term of the sequence of non-zero natural numbers (a_n)n≥1, with the property that gcd(a_m, a_n, a_p) = gcd(m^2 ,n^2 ,p^2), for any distinct non-zero natural numbers m, n, p.

⁡Note that gcd(a,b,c) denotes the greatest common divisor of the natural numbers a,b,c .
1 reply
BBNoDollar
Today at 5:53 PM
BBNoDollar
an hour ago
No more topics!
Polynomial mapping set of divisors to set of divisors
eduD_looC   11
N Apr 26, 2025 by Ilikeminecraft
Source: CMO 2023 P4
Let $f(x)$ be a non-constant polynomial with integer coefficients such that $f(1) \neq 1$. For a positive integer $n$, define $\text{divs}(n)$ to be the set of positive divisors of $n$.

A positive integer $m$ is $f$-cool if there exists a positive integer $n$ for which $$f[\text{divs}(m)]=\text{divs}(n).$$Prove that for any such $f$, there are finitely many $f$-cool integers.

(The notation $f[S]$ for some set $S$ denotes the set $\{f(s):s \in S\}$.)
11 replies
eduD_looC
Mar 11, 2023
Ilikeminecraft
Apr 26, 2025
Polynomial mapping set of divisors to set of divisors
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G H BBookmark kLocked kLocked NReply
Source: CMO 2023 P4
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eduD_looC
6610 posts
#1 • 1 Y
Y by Rounak_iitr
Let $f(x)$ be a non-constant polynomial with integer coefficients such that $f(1) \neq 1$. For a positive integer $n$, define $\text{divs}(n)$ to be the set of positive divisors of $n$.

A positive integer $m$ is $f$-cool if there exists a positive integer $n$ for which $$f[\text{divs}(m)]=\text{divs}(n).$$Prove that for any such $f$, there are finitely many $f$-cool integers.

(The notation $f[S]$ for some set $S$ denotes the set $\{f(s):s \in S\}$.)
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jj_ca888
2726 posts
#2 • 1 Y
Y by Ru83n05
Take prime $p$ dividing $f(1)$ and $m >> N$ where $f$ is strictly increasing past $N$. $f$ of largest factors of $m$ must correspond to largest factors of $n$, and size argument (relying on the fact that $l^n$ cannot equal $p$ ever for n > 1) wins
This post has been edited 2 times. Last edited by jj_ca888, Mar 11, 2023, 1:49 AM
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js104
14 posts
#3 • 1 Y
Y by A64298347
alternatively: there are finitely many $a$ where $f(a)=1$ (at most degree of $f$), so one must divide infinitely many $m$, so then $f(m/a)\mid f(m) \implies f(x)\mid f(ax)$ has infinitely many solutions (again for large enough $m$ where it's strictly increasing and $f(m)>f(n)$ for all $0<n<m$). do division algorithm, $f(ax)=qf(x)+g(x)$ where deg g < deg f, so $f(x)\mid g(x)$ infinitely many times. after some point $|f(x)| > |g(x)|$ so $g(x)=0$ for infinitely many $x$ and has to be the zero polynomial; $f(ax)=qf(x) \implies f(x)$ is a monomial $cx^d$, but this is clearly bad (no integer sols to $f(a)=1$)
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L567
1184 posts
#4 • 4 Y
Y by VicKmath7, mathscrazy, Ru83n05, jmiao
Note that eventually the function will be strictly increasing, therefore we must have $f(m) = n$ and $f(\frac{m}{p}) = \frac{n}{q}$, where $p$ and $q$ are the smallest prime divisors of $m$ and $n$ respectively. These two can be rewritten to $$f(\frac{m}{p}) = \frac{f(m)}{q}$$But note that $f(1) \neq 1$ be among the divisors of $n$ and hence divide $n$, so the value of $q$ is actually bounded. As a consequence, due to the above equation, $p$ is bounded too, therefore there exist infinitely many values of $m$ which have the same value of $p$, and for all of these we have that $f(k) \mid f(kp)$ for $k = \frac{m}{p}$. Since this is true for infinitely many big integers, it forces that $f(kp) = Q(k)f(k)$, implying $f$ can only be a monomial. But this is clearly impossible since nothing goes to one here except one itself.
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IAmTheHazard
5001 posts
#5 • 1 Y
Y by centslordm
Some people might get a -1 by not considering $f(1)<0$ (in particular $f(1)=-1$)?

We will prove that there are finitely many cool pairs of positive integers $(m,n)$ such that $f(\text{divs}(m))=\text{divs}(n)$. Throwing out the finitely many pairs with $m=1$ or $n=1$, The key is to consider the least prime divisor of $m$ and $n$. If $f(1)\leq 0$ we cannot have any $f$-cool pairs because $f(\text{divs}(m))$ will always have a nonpositive number but $\text{divs}(n)$ cannot. Thus $f(1)>1$, so since $f(1)$ must divide $n$ the least prime divisor of $n$ in any cool pair $(m,n)$ is bounded. Furthermore, since $1 \in \text{divs}(n)$, one of the divisors $d$ of $m$ must have $f(d)=1$. Since there are finitely many solutions in the positive integers to this, and $1$ is not one of them, the least prime divisor of any $m$ in a cool pair $(m,n)$ is also bounded (in particular, by the largest prime divisor over all these solutions).

For $m$ large enough, the largest value in $f(\text{divs}(m))$ should be $f(m)$, and the second-largest should be $f(m/p)$, where $p$ is the least prime divisor of $m$. Thus for a good pair $(m,n)$ we should have $(m,n)$, we should have $f(m)=n$ and $f(m/p)=n/q$ where $q$ is the least prime divisor of $n$. Since we proved $p$ and $q$ were bounded, if there are infinitely many cool pairs $(m,n)$, by pigeonhole there should be some pair of primes $(p,q)$ that occurs infinitely often, i.e. there are infinitely many $m$ with
$$f\left(\frac{m}{p}\right)=\frac{f(m)}{q}.$$Thus this must be a polynomial equality. But if there is an $x^k$ term in $f$ with nonzero coefficient, it gets multiplied by $p^{-k}$ on the LHS and $q^{-1}$ on the RHS, implying that $k$ must be unique, so $f(x)=cx^d$ is the only polynomial we must consider, where $c>1$. However, such polynomials will not have any integers solutions to $f(x)=1$ by taking modulo $c$, which is necessary for any pair $(m,n)$ to be cool, so we are done. $\blacksquare$
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Nuterrow
254 posts
#6
Y by
Sad problem. Assume contradiction and notice that there are finitely many $a_i$ such that $f(a_i)=1$ and suppose $\mathbb{P}$ is the prime-set of these numbers $a_i$. Let $1 < d_a < d_{a-1} < \cdots < d_1 <m$ and $1 < e_a < e_{a-1} < \cdots < e_1 <n$ be the divisors of $m$ and $n$ respectively where $m$ is cool and $n$ is its cool conjugate. WLOG assume that the leading coefficient of $f$ is positive and we can find $N$ such that for all $m \geq N$, $f(m)=n$ and $f(d_i)=e_i$ where $1 \leq i \leq a$. This means, $$f(d_a) \mid f(m)=f(d_ad_1)$$Notice that $d_a$ is bounded as $d_a \in \mathbb{P}$, thus for some prime $d_a = p_i \in \mathbb{P}$, the divisibility is true infinitely many times. Now we can force $f(d_1x) = f(x)g(x)$ for some polynomial $g$ for all $x$ (say by the euclidean algorithm), thus $f(x) = cx^d$ but then $f$ is never mapped to $1$ which is a contradiction. $\blacksquare$
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Bacteria
256 posts
#7 • 2 Y
Y by jmiao, khina
The original (harder) version of this problem had the condition that $f(x)\neq x$ (as polynomials) instead of $f(1)\neq 1$, with the goal to prove that there are finitely many $f-$cool composite integers.
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duotran
7 posts
#8
Y by
Bacteria wrote:
The original (harder) version of this problem had the condition that $f(x)\neq x$ (as polynomials) instead of $f(1)\neq 1$, with the goal to prove that there are finitely many $f-$cool composite integers.

May I ask which problem you referred to above or in which competition?
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HamstPan38825
8857 posts
#9
Y by
Suppose WLOG that the leading coefficient of $f$ is positive. Let $M$ be the largest local maximum of $f$, and let $N$ be the largest number such that $f(N) = M$. Assume that there exist infinitely many $f$-cool integers $m$; then there exists such a $m$ with $m > N$.

Let $p_1$ be the smallest prime factor of $m$ and $q_1$ the smallest factor of $n$. It follows that $f(n) = m > M$ and $f\left(\frac n{p_1}\right) = \frac m{p_1}$. But setting $c$ to be the leading coefficient of $f$, by taking $n$ sufficiently large, we can force $$(p_1^d - 1) f\left(\frac n{p_1}\right) < f(n) < (p_1^d + 1) f\left(\frac n{p_1}\right).$$In particular, this implies that $q_1 = p_1^d$, which is absurd for $d \geq 2$.

Thus $d=1$; but because there exists an integer $k$ with $f(k) = 1$, it follows the leading coefficient of $f$ is $\pm 1$. Having $f(1) < 0$, on the other hand, is clearly impossible, so this yields an immediate contradiction.
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L13832
266 posts
#10
Y by
Impossible to write the solution correctly

FTSOC assume there there are infinitely many $f-$cool integers.
Claim I: $f(m)=n$
Proof: It is easy to see that $f$ has positive coefficients. So $f(m)$ and $n$ are the largest value in the sets $f[d(m)]$ and $d(n)$ respectively , so $f(m)=n$.

Also note that $1$ is the smallest $d(m)$, so let $f(k)=1$ for some $k$. Also $k\neq 1$ according to the problem.
Claim II: $f(k)\mid f(ak)$, for $f(k)=1$
Proof: $\exists k$ such that $\displaystyle f\biggl(\frac{m}{k}\biggr) \in f[d(m)]=d(f(m))\implies f\biggl(\frac{m}{k}\biggr)\bigg|f(m)\implies f(k)\mid f(ak)$ which has infinite solutions.

Let $\text{deg}(f)=d$ and $f(ak)=g\cdot f(k)+h(k) \implies f(k)\mid h(k)$, but $\text{deg}(h)<\text{deg}(f)$ hence we have a contradiction, so $h$ is $0$. We get $f(ak)=g \cdot f(k) \implies f(k)=ck^d$. But if $c=1$ then $f(1)=1$ which is a contradiction so there are finitely $f-$cool integers. $\blacksquare$
This post has been edited 1 time. Last edited by L13832, Sep 9, 2024, 4:18 PM
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SomeonesPenguin
128 posts
#11 • 1 Y
Y by zzSpartan
Notice that since $f$ is eventually positive, there exists $N$ such that for every $u \ge N$ and $u > v$ we have $f(u) > f(v)$. Hence if $m$ is big enough we get that $f(m)$ is the biggest so $f(m) = n$. We also exclude prime $m$ because we can only have finitely many of these. Now let $p$ be the smallest prime dividing $m$ and $q$ be the smallest prime dividing $n$. Since $m$ isn't prime, we have that $\frac{m}{p} \ge p$ so it can be sufficiently large. So we get that $f\left(\frac{m}{p}\right)$ is the second biggest and so $f\left(\frac{m}{p}\right)=\frac{n}{q}$. So we finally get: $$f\left(\frac{m}{p}\right)=\frac{f(m)}{q}$$
Also notice that $f(1)$ must be greater than $1$ hence we have that $f(1)\ge q$ since $f(1)\in \text{divs}(n)$. So we get that $$f\left(\frac{m}{p}\right)\ge \frac{f(m)}{f(1)}$$
Therefore $p \le f(1)$ (for sufficiently large $m$). And so $p$ is bounded meaning that there are infinitely many $f$-cool integers with least prime factor $p$. Hence we must have $f(x)=P(x)f\left(\frac{x}{p}\right)$. Hence $\text{deg}P = 0$ and looking at some non-zero coefficient of $x^{k}$ we get $P(x) = 1$ which is clearly a contradiction.
This post has been edited 1 time. Last edited by SomeonesPenguin, Aug 9, 2024, 11:08 AM
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Ilikeminecraft
612 posts
#12
Y by
Obviously note that the leading coefficient of $f$ must be positive. Clearly, there are finitely many values of $f$ such that $f(a) = 1.$ Hence, there has to exist a value $a$ so that there are infinitely many $k$ such that $ak$ is $f$-cool. Take an arbitrarily large value of $k$ such that $f$ is increasing over this interval. Hence, $f(ak)$ is the largest value in the set in the left hand side, which implies $m = f(ak).$ Thus, $f(k)\mid f(ak)$ for infinitely many values $k$. Let $d = \operatorname{deg} f$. Suppose $f(ax) = a^d f(x) + g(x),$ where $g \in \mathbb Z[x]$ by division algorithm. Clearly, $\operatorname{deg}g < \operatorname{deg}f.$ However, we also have $f(x)\mid g(x)$ for infinitely many values of $x.$ This implies that $g \equiv 0.$ Thus, $f(x) = ax^d,$ which is impossible to achieve $f(x) = 1.$
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