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Bounds on degree of polynomials
Phorphyrion   3
N 8 minutes ago by Kingsbane2139
Source: 2020 Israel Olympic Revenge P3
For each positive integer $n$, define $f(n)$ to be the least positive integer for which the following holds:

For any partition of $\{1,2,\dots, n\}$ into $k>1$ disjoint subsets $A_1, \dots, A_k$, all of the same size, let $P_i(x)=\prod_{a\in A_i}(x-a)$. Then there exist $i\neq j$ for which
\[\deg(P_i(x)-P_j(x))\geq \frac{n}{k}-f(n)\]
a) Prove that there is a constant $c$ so that $f(n)\le c\cdot \sqrt{n}$ for all $n$.

b) Prove that for infinitely many $n$, one has $f(n)\ge \ln(n)$.
3 replies
Phorphyrion
Jun 11, 2022
Kingsbane2139
8 minutes ago
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Number of roots of boundary preserving unit disk maps
Assassino9931   3
N Yesterday at 2:12 AM by bsf714
Source: Vojtech Jarnik IMC 2025, Category II, P4
Let $D = \{z\in \mathbb{C}: |z| < 1\}$ be the open unit disk in the complex plane and let $f : D \to D$ be a holomorphic function such that $\lim_{|z|\to 1}|f(z)| = 1$. Let the Taylor series of $f$ be $f(z) = \sum_{n=0}^{\infty} a_nz^n$. Prove that the number of zeroes of $f$ (counted with multiplicities) equals $\sum_{n=0}^{\infty} n|a_n|^2$.
3 replies
Assassino9931
May 2, 2025
bsf714
Yesterday at 2:12 AM
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|A/pA|<=p, finite index=> isomorphism - OIMU 2008 Problem 7
Jorge Miranda   2
N Thursday at 8:00 PM by pi_quadrat_sechstel
Let $A$ be an abelian additive group such that all nonzero elements have infinite order and for each prime number $p$ we have the inequality $|A/pA|\leq p$, where $pA = \{pa |a \in A\}$, $pa = a+a+\cdots+a$ (where the sum has $p$ summands) and $|A/pA|$ is the order of the quotient group $A/pA$ (the index of the subgroup $pA$).

Prove that each subgroup of $A$ of finite index is isomorphic to $A$.
2 replies
Jorge Miranda
Aug 28, 2010
pi_quadrat_sechstel
Thursday at 8:00 PM
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Prove the statement
Butterfly   8
N Thursday at 7:32 PM by oty
Given an infinite sequence $\{x_n\} \subseteq [0,1]$, there exists some constant $C$, for any $r>0$, among the sequence $x_n$ and $x_m$ could be chosen to satisfy $|n-m|\ge r $ and $|x_n-x_m|<\frac{C}{|n-m|}$.
8 replies
Butterfly
May 7, 2025
oty
Thursday at 7:32 PM
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Functional equation from limit
IsicleFlow   1
N Thursday at 4:22 PM by jasperE3
Is there a solution to the functional equation $f(x)=\frac{1}{1-x}f(\frac{2 \sqrt{x} }{1-x}), f(0)=1$ Such That $ f(x) $ is even?
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1 reply
IsicleFlow
Jun 9, 2024
jasperE3
Thursday at 4:22 PM
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f(m+n)≤f(m)f(n) implies existence of limit
Etkan   2
N Thursday at 3:19 PM by Etkan
Let $f:\mathbb{Z}_{\geq 0}\to \mathbb{Z}_{\geq 0}$ satisfy $f(m+n)\leq f(m)f(n)$ for all $m,n\in \mathbb{Z}_{\geq 0}$. Prove that$$\lim \limits _{n\to \infty}f(n)^{1/n}=\inf \limits _{n\in \mathbb{Z}_{>0}}f(n)^{1/n}.$$
2 replies
Etkan
May 15, 2025
Etkan
Thursday at 3:19 PM
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Collinearity in a Harmonic Configuration from a Cyclic Quadrilateral
kieusuong   0
Thursday at 2:26 PM
Let \((O)\) be a fixed circle, and let \(P\) be a point outside \((O)\) such that \(PO > 2r\). A variable line through \(P\) intersects the circle \((O)\) at two points \(M\) and \(N\), such that the quadrilateral \(ANMB\) is cyclic, where \(A, B\) are fixed points on the circle.

Define the following:
- \(G = AM \cap BN\),
- \(T = AN \cap BM\),
- \(PJ\) is the tangent from \(P\) to the circle \((O)\), and \(J\) is the point of tangency.

**Problem:**
Prove that for all such configurations:
1. The points \(T\), \(G\), and \(J\) are collinear.
2. The line \(TG\) is perpendicular to chord \(AB\).
3. As the line through \(P\) varies, the point \(G\) traces a fixed straight line, which is parallel to the isogonal conjugate axis (the so-called *isotropic line*) of the centers \(O\) and \(P\).

---

### Outline of a Synthetic Proof:

**1. Harmonic Configuration:**
- Since \(A, N, M, B\) lie on a circle, their cross-ratio is harmonic:
\[ (ANMB) = -1. \]- The intersection points \(G = AM \cap BN\), and \(T = AN \cap BM\) form a well-known harmonic setup along the diagonals of the quadrilateral.

**2. Collinearity of \(T\), \(G\), \(J\):**
- The line \(PJ\) is tangent to \((O)\), and due to harmonicity and projective duality, the polar of \(G\) passes through \(J\).
- Thus, \(T\), \(G\), and \(J\) must lie on a common line.

**3. Perpendicularity:**
- Since \(PJ\) is tangent at \(J\) and \(AB\) is a chord, the angle between \(PJ\) and chord \(AB\) is right.
- Therefore, line \(TG\) is perpendicular to \(AB\).

**4. Quasi-directrix of \(G\):**
- As the line through \(P\) varies, the point \(G = AM \cap BN\) moves.
- However, all such points \(G\) lie on a fixed line, which is perpendicular to \(PO\), and is parallel to the isogonal (or isotropic) line determined by the centers \(O\) and \(P\).

---

**Further Questions for Discussion:**
- Can this configuration be extended to other conics, such as ellipses?
- Is there a pure projective geometry interpretation (perhaps using polar reciprocity)?
- What is the locus of point \(T\), or of line \(TG\), as \(P\) varies?

*This configuration arose from a geometric investigation involving cyclic quadrilaterals and harmonic bundles. Any insights, counterexamples, or improvements are warmly welcomed.*
0 replies
kieusuong
Thursday at 2:26 PM
0 replies
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Find solution of IVP
neerajbhauryal   2
N Thursday at 1:50 PM by Mathzeus1024
Show that the initial value problem \[y''+by'+cy=g(t)\] with $y(t_o)=0=y'(t_o)$, where $b,c$ are constants has the form \[y(t)=\int^{t}_{t_0}K(t-s)g(s)ds\,\]

What I did
2 replies
neerajbhauryal
Sep 23, 2014
Mathzeus1024
Thursday at 1:50 PM
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fourier series?
keroro902   2
N Thursday at 12:54 PM by Mathzeus1024
f(x)=$\sum _{n=0}^{\infty } \text{cos}(nx)/2^{n}$
f(x) = ?
thanks
2 replies
keroro902
May 14, 2010
Mathzeus1024
Thursday at 12:54 PM
J
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Sets on which a continuous function exists
Creativename27   1
N May 15, 2025 by alexheinis
Source: My head
Find all $X\subseteq R$ that exist function $f:R\to R$ such $f$ continuous on $X$ and discontinuous on $R/X$
1 reply
Creativename27
May 15, 2025
alexheinis
May 15, 2025
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Japanese Olympiad
parkjungmin   6
N May 15, 2025 by mathNcheese_aops
It's about the Japanese Olympiad

I can't solve it no matter how much I think about it.

If there are people who are good at math

Please help me.
6 replies
parkjungmin
May 10, 2025
mathNcheese_aops
May 15, 2025
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J
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Diophantine equation !
ComplexPhi   6
N Apr 22, 2025 by Mr.Sharkman
Source: Romania JBMO TST 2015 Day 1 Problem 4
Solve in nonnegative integers the following equation :
$$21^x+4^y=z^2$$
6 replies
ComplexPhi
May 14, 2015
Mr.Sharkman
Apr 22, 2025
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Diophantine equation !
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Source: Romania JBMO TST 2015 Day 1 Problem 4
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ComplexPhi
455 posts
ComplexPhi
#1 h  May 14, 2015, 6:52 AM • 2 Y
Y by ahmedosama, Adventure10
Solve in nonnegative integers the following equation :
$$21^x+4^y=z^2$$
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BSJL
641 posts
BSJL
#2 h  May 14, 2015, 7:10 AM • 2 Y
Y by Adventure10, Mango247
Rewrite the condition:

\[ 3^x \cdot 7^x=(z-2^y)(z+2^y) \]

Since $ z-2^y, z+2^y $ can't be the multiples of 3,7 at the some time, so we have two cases.

Case: $ \boxed{z-2^y=1,z+2^y=21^x} $

We have $ 2^{y+1}=21^x-1 $. Notice that $ 5|(21^x-1) $, thus, there isn't any solution.

Case: $ \boxed{z-2^y=3^x,z+2^y=7^x} $

In this case, we have $ 2^{y+1}=7^x-3^x $, which can be solve immediately by Zsigmody.

The only solution is $ x=1,y=1 \rightarrow \{x,y,z\}=\{1,1,5\}$
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randomasdf97
99 posts
randomasdf97
#3 h  May 14, 2015, 9:38 AM • 1 Y
Y by Adventure10
BSJL wrote:
In this case, we have $ 2^{y+1}=7^x-3^x $, which can be solve immediately by Zsigmody.

The only solution is $ x=1,y=1 \rightarrow \{x,y,z\}=\{1,1,5\}$

By Zsigmondy, more generally $a^n+b^n$ has at least two prime divisors when $a>b>0, n\ge 2, (a,b,n)\neq (2,1,3)$.

$a^n - b^n$ has at least two prime divisors when $a>b+1>1, n\ge 2, (a,b)=1, (a,b,n)\neq (a, 2^m-a, 2)$ for any integer $m$.

Here $7>3+1>1, (7,3)=1, (a,b,n)\neq (a,2^m-a,2)$ for any integer $m$, since $x\neq 2$ by checking.

$x\ge 2$ gives contradiction by above Zsigmondy lemma, so $x\in\{0,1\}$. $x\neq 0$. $x=1$ gives $(x,y,z)=(1,1,5)$.

Don't confuse set notation. It is false that $\{x,y,z\}=\{1,1,5\}$ is a solution (clearly $z\neq 1$).

Zsigmondy is overkill here (the proof of Zs isn't elementary). $x$ even gives $2^{y+1}=(7^{x/2}-3^{x/2})(7^{x/2}+3^{x/2})$.

$7^{x/2}+3^{x/2}\equiv 2\pmod {4}$, so $7^{x/2}+3^{x/2}=2$ and $x=0$, contr.

So $x$ is odd. mod $8$ gives $7^x-3^x\equiv 4\pmod {8}$, so $y=1$, and $(x,y,z)=(1,1,5)$.
This post has been edited 7 times. Last edited by randomasdf97, May 14, 2015, 8:09 PM
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neverlose
117 posts
neverlose
#4 h  Jun 1, 2015, 2:31 PM • 2 Y
Y by Adventure10, Mango247
ComplexPhi wrote:
Solve in nonnegative integers the following equation :
$$21^x+4^y=z^2$$

If $x = 0$ we get $(z - 1)(z + 1) = 4^y$,if we look at equation $modulo$ $4$ we obtain,if $z-1 \equiv 0($ $mod$ $4)$
$\implies$ $z+1 \equiv 0$ $(mod$ $4)$,and similarly if $z - 1 \equiv 0$ $(mod$ $4)$ we obtain contradiction so $x \neq 0$.
If $y = 0$ we get $(z - 1)(z + 1) = 21^x$,$z-1\neq 1$ and $z+1\neq 1$.
If both $z-1$,$z+1$ are divisible with 21 we obtain that $gcd(z-1,z+1)\leq 2$,contradition.
If $z - 1 = 3^x$,$z+1 = 7^x$ $\implies$ $7^x - 3^x = 2$,contradiction $7^x - 3^x$ is divisible with 4.
If $z - 1 = 7^x$,$z+1 = 3^x$ $\implies$ $7^x < 3^x$,contradiction.
We obtain that $y\neq 0$.
$3^x\cdot7^x = (z - 2^y)(z+2^y)$.
If $z - 2^y = 1$ $\implies$ $z = 2^y + 1$ and $4^y+2^{y+1}+1=4^y+21^x$ or $21^x-1^x = 2^{y+1}$,but $21^x - 1^x$ if divisible with $20$,contradiction.
If $z +2^y = 1$ $\implies$ $y = 0$,contradiction.
If both $z - 2^y$ and $z + 2^y$ are divisible with 21 $\implies$ $21\equiv 0$ $(mod$ $2)$,contradiction.
If $z - 2^y = 3^x$ and $z + 2^y = 7^x$ $\implies$ $3^x + 2^{y+1} = 7^x$.
If $y\ge 2$ $\implies$ $3^x\equiv 0$ $(mod$ $8)$ but $3^x\equiv 3,1$ $(mod$ $8)$ and $7^x\equiv -1,1$ $(mod$ $8)$,contradiction $\implies$ $y = 1$,$x = 1$ and $z = 5$.
If $z - 2^y = 7^x$ and $z + 2^y = 3^x$ $\implies$ $3^x > 7^x$,contradiction.
So the single solution is $(x,y,z) = (1,1,5)$.
This post has been edited 6 times. Last edited by neverlose, Feb 8, 2016, 4:43 PM
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Assassino9931
1354 posts
Assassino9931
#5 h  Sep 25, 2023, 2:46 PM
Y by
To fill in, here is an elementary way to solve $2^{y+1} = 7^x - 3^x$ for $y\geq 2$. By mod $8$ we get that $x$ is even and so $2^{y+1} = (7^{x/2} - 3^{x/2})(7^{x/2} + 3^{x/2})$, thus (as the two factors have gcd $2$) $7^{x/2} - 3^{x/2} = 2$, which is impossible since $3^{x/2} \geq 9$ and $(7/3)^{x/2} - 1 > 5$ for $x\geq 4$.
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aops.c.c.
1 post
aops.c.c.
#11 h  Apr 22, 2025, 6:58 AM • 1 Y
Y by Turkish_sniper
If $y=0 \implies 3^x.7^x=(z-1)(z+1)$
By Euclidean Algorithm, there is 2 cases.
Case 1: $z+1=7^x$
$z-1=3^x$
So $7^x-3^x=2$ which is impossible.
Case 2 $z-1=1$
$z+1=21^x$ is a contradiction.

If $y=1 \implies 3^x.7^x=(z-2).(z+2)$
By Euclidean Algorithm, again there is 2 cases.
Case 1: $z-2=3^x$
$z+2=7^x$
$\implies 7^x-3^x=4$
Clearly there is only 1 solution. $\boxed{(x,y,z)=(1,1,5)}$
Case 2: $z-2=1$
$z+2=21^x$ is impossible.

If $y>1$
By $mod 8$ x is even.
So the new equation is $2^{2y}=(z-21^{x/2}).(z+21^{x/2})$
By Euclidean Algorithm $z-21^{x/2}=2$ $z+21^{x/2}=2^{2y-1}$
$\implies 2.21^{x/2}=2^{2y-1}-2$
If we divide both sides by 2 $21^{x/2}=2^{2y-2}-1$ And this is also a contradiction by $mod 8$
So the only solution is $\boxed{(x,y,z)=(1,1,5)}$
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Mr.Sharkman
500 posts
Mr.Sharkman
#12 h  Apr 22, 2025, 1:52 PM
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The answer is $\boxed{(1,1, 5)}$ only. We have that $z$ is odd, so $(z^{2}-2^{y})(z^{2}+2^{y}) = 3^{x}*7^{x},$ and thus $2^{y+1} = 7^{x}-3^{x}.$ Now, $\nu_{2}(7^{x}-3^{x}) = 3+\nu_{2}(x)-1 = \nu_{2}(x)+2,$ so $2^{\nu_{2}(x)+2} = 7^{x}-3^{x} \le 4x,$ which is true only when $x \le 1,$ so our solution is $x=1$, $y=1$, and then $z=5.$
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