Multiplicative polynomial exactly 2025 times

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0 replies

jlacosta

May 1, 2025

0 replies

Help me this problem. Thank you

illybest 3

N 13 minutes ago by jasperE3

Find f: R->R such that
f( xy + f(z) ) = (( xf(y) + yf(x) )/2) + z

3 replies

illybest

Today at 11:05 AM

jasperE3

13 minutes ago

line JK of intersection points of 2 lines passes through the midpoint of BC

parmenides51 4

N 31 minutes ago by reni_wee

Source: Rioplatense Olympiad 2018 level 3 p4

Let $ABC$ be an acute triangle with $AC> AB$ . be $\Gamma$ the circumcircle circumscribed to the triangle $ABC$ and $D$ the midpoint of the smallest arc $BC$ of this circle. Let $E$ and $F$ points of the segments $AB$ and $AC$ respectively such that $AE = AF$ . Let $P \neq A$ be the second intersection point of the circumcircle circumscribed to $AEF$ with $\Gamma$ . Let $G$ and $H$ be the intersections of lines $PE$ and $PF$ with $\Gamma$ other than $P$ , respectively. Let $J$ and $K$ be the intersection points of lines $DG$ and $DH$ with lines $AB$ and $AC$ respectively. Show that the $JK$ line passes through the midpoint of $BC$

4 replies

parmenides51

Dec 11, 2018

reni_wee

31 minutes ago

AGM Problem(Turkey JBMO TST 2025)

HeshTarg 3

N 32 minutes ago by ehuseyinyigit

Source: Turkey JBMO TST Problem 6

Given that $x, y, z > 1$ are real numbers, find the smallest possible value of the expression:
$\frac{x^3 + 1}{(y-1)(z+1)} + \frac{y^3 + 1}{(z-1)(x+1)} + \frac{z^3 + 1}{(x-1)(y+1)}$

3 replies

HeshTarg

an hour ago

ehuseyinyigit

32 minutes ago

Shortest number theory you might've seen in your life

AlperenINAN 8

N an hour ago by HeshTarg

Source: Turkey JBMO TST 2025 P4

Let $p$ and $q$ be prime numbers. Prove that if $pq(p+1)(q+1)+1$ is a perfect square, then $pq + 1$ is also a perfect square.

8 replies

AlperenINAN

Yesterday at 7:51 PM

HeshTarg

an hour ago

No more topics!

Multiplicative polynomial exactly 2025 times

Assassino9931 1

N Apr 10, 2025 by sami1618

Source: Bulgaria Balkan MO TST 2025

Does there exist a polynomial $P$ on one variable with real coefficients such that the equation $P(xy) = P(x)P(y)$ has exactly $2025$ ordered pairs $(x,y)$ as solutions?

1 reply

Assassino9931

Apr 9, 2025

sami1618

Apr 10, 2025

Multiplicative polynomial exactly 2025 times

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Source: Bulgaria Balkan MO TST 2025

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Assassino9931

1347 posts

Assassino9931

#1 h Apr 9, 2025, 10:14 PM

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Does there exist a polynomial $P$ on one variable with real coefficients such that the equation $P(xy) = P(x)P(y)$ has exactly $2025$ ordered pairs $(x,y)$ as solutions?

This post has been edited 1 time. Last edited by Assassino9931, Apr 9, 2025, 10:34 PM

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sami1618

909 posts

sami1618

#2 h Apr 10, 2025, 7:43 PM • 3 Y

Y by Assassino9931, kiyoras_2001, Mysteriouxxx

this should work

Answer: Yes, such a polynomial exists.
Solution. Let $\alpha_1,\alpha_2,\dots,\alpha_n$ be distinct non-zero real numbers and consider the polynomial $P(z)=-\prod_{i=1}^n(z-\alpha_i)^2$ Since $P\leq 0$ for all real numbers, $P(xy)\leq 0\leq P(x)P(y)$ . Thus we must have $P(xy)=0=P(x)P(y)$ . In other words $xy$ and at lest one of $x$ and $y$ must lie in $\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ . Let $N$ be the number of ordered integer triplets $(i,j,k)$ , with $1\leq i,j,k\leq n$ , such that $\alpha_i\alpha_j=\alpha_k$ .

The number of ordered pairs $(x,y)$ where both $x$ and $xy$ lie in $\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ is $n^2$ . Similarly, the number of ordered pairs if $y$ and $xy$ lie in $\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ is $n^2$ . The number of ordered pairs where $x$ , $y$ , and $xy$ all lie in $\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ is equal to $N$ . Thus the total number of pairs that satisfy the equation is $2n^2-N$ .

Now it suffices to exhibit a set $\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ where $2n^2-N=2025$ . Consider the set $\{2^{16},2^{17},\dots,2^{48}\}$ . It is not hard to see that $n=33$ and $N=153$ giving $2n^2-N=2025$ as required.

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