Function on positive integers with two inputs

Find all positive integers $n\geq1$ such that there exists a pair $(a,b)$ of positive integers, such that $a^2+b+3$ is not divisible by the cube of any prime, and $n=\frac{ab+3b+8}{a^2+b+3}.$

59 replies

DottedCaculator

Jul 12, 2022

SimplisticFormulas

44 minutes ago

Interesting inequalities

sqing 2

N an hour ago by sqing

Source: Own

Let $a,b,c\geq 0 ,2a +ab +abc \geq 9.$ Prove that
$a+b+c \geq 4$ $a+b+\frac{1}{4}c \geq \frac{13}{4}$ Let $a,b,c\geq 0 ,2a +ab +4abc \geq 9.$ Prove that
$a+b+c+abc \geq 4$ $a+b+4c \geq 4$ $a+b+c \geq \frac{13}{4}$ $a+\frac{3}{2}b+4c \geq 3(\sqrt{6}-1)$ $a+\frac{9}{4}b+4c \geq \frac{9}{2}$ $a+\frac{4}{3}b+4c \geq 4\sqrt 3-\frac{8}{3}$

2 replies

sqing

2 hours ago

sqing

an hour ago

IMO Shortlist Problems

ABCD1728 3

N an hour ago by Phorphyrion

Source: IMO official website

Where can I get the official solution for ISL before 2005? The official website only has solutions after 2006. Thanks :)

3 replies

ABCD1728

Yesterday at 12:44 PM

Phorphyrion

an hour ago

A convex pentagon has rational sides and equal angles

Valentin Vornicu 2

N 2 hours ago by Qing-Cloud

Source: Balkan MO 2001, problem 2

A convex pentagon $ABCDE$ has rational sides and equal angles. Show that it is regular.

2 replies

Valentin Vornicu

Apr 24, 2006

Qing-Cloud

2 hours ago

Hard diophant equation

MuradSafarli 3

N 2 hours ago by MuradSafarli

Find all positive integers $x, y, z, t$ such that the equation

$2017^x + 6^y + 2^z = 2025^t$
is satisfied.

3 replies

MuradSafarli

Yesterday at 6:12 PM

MuradSafarli

2 hours ago

< NA'T = < ADT wanted, starting with a right triangle, symmetric, projections

parmenides51 3

N 3 hours ago by tilya_TASh

Source: JBMO Shortlist 2018 G2

Let $ABC$ be a right angled triangle with $\angle A = 90^o$ and $AD$ its altitude. We draw parallel lines from $D$ to the vertical sides of the triangle and we call $E, Z$ their points of intersection with $AB$ and $AC$ respectively. The parallel line from $C$ to $EZ$ intersects the line $AB$ at the point $N$ . Let $A'$ be the symmetric of $A$ with respect to the line $EZ$ and $I, K$ the projections of $A'$ onto $AB$ and $AC$ respectively. If $T$ is the point of intersection of the lines $IK$ and $DE$ , prove that $\angle NA'T = \angle ADT$ .

3 replies

parmenides51

Jul 22, 2019

tilya_TASh

3 hours ago

x(x - y) = 8y - 7 in NxN

parmenides51 4

N 3 hours ago by Namisgood

Source: JBMO 2008 Shortlist N1

Find all the positive integers $x$ and $y$ that satisfy the equation $x(x - y) = 8y - 7$

4 replies

parmenides51

Oct 14, 2017

Namisgood

3 hours ago

Insects walk

Giahuytls2326 0

3 hours ago

Source: somewhere in the internet

A 100 × 100 chessboard is divided into unit squares. Each square has an arrow pointing up, down, left, or right. The board square is surrounded by a wall, except to the right of the top right corner square. An insect is placed in one of the squares.

Every second, the insect moves one unit in the direction of the arrow in its square. As the insect moves, the arrow of the square it just left rotates 90° clockwise.

If the specified movement cannot be performed, then the insect will not move for that second, but the arrow in the square it is standing on will still rotate. Is it possible that the insect never leaves the board?

0 replies

Giahuytls2326

3 hours ago

0 replies

Equal sum of digits

Fudicuehfosonrcjeong 0

4 hours ago

Is it true that for any two positive integers a, b there exists a positive integer k such that s(ka)=s(kb), where s(n) is sum of digits in base 10?

0 replies

Fudicuehfosonrcjeong

4 hours ago

0 replies

Common tangent of mixtilinear incircles

CyclicISLscelesTrapezoid 3

N 4 hours ago by Ilikeminecraft

Source: MOP 2020/1Z

Let $ABCD$ be a quadrilateral inscribed in circle $\Omega$ . Circles $\omega_A$ and $\omega_D$ are drawn internally tangent to $\Omega$ , such that $\omega_A$ is tangent to $\overline{AB}$ and $\overline{AC}$ while $\omega_D$ is tangent to $\overline{DB}$ and $\overline{DC}$ . Prove that we can draw a line parallel to $\overline{AD}$ which is simultaneously tangent to both $\omega_A$ and $\omega_D$ .

3 replies

CyclicISLscelesTrapezoid

Jan 6, 2023

Ilikeminecraft

4 hours ago

Function on positive integers with two inputs

Assassino9931 2

N Apr 23, 2025 by Assassino9931

Source: Bulgaria Winter Competition 2025 Problem 10.4

The function $f: \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ is such that $f(a,b) + f(b,c) = f(ac, b^2) + 1$ for any positive integers $a,b,c$ . Assume there exists a positive integer $n$ such that $f(n, m) \leq f(n, m + 1)$ for all positive integers $m$ . Determine all possible values of $f(2025, 2025)$ .

2 replies

Assassino9931

Jan 27, 2025

Assassino9931

Apr 23, 2025

Function on positive integers with two inputs

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Source: Bulgaria Winter Competition 2025 Problem 10.4

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Assassino9931

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Assassino9931

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Assassino9931

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#3 h Apr 23, 2025, 4:01 PM

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Official Solution

The function $f \equiv 1$ satisfies the condition and $f(2025, 2025) = 1$ . Suppose that it is not possible $f(2025, 2025) > 1$ . Comparing $(a, b, c) = (x, y, y)$ and $(a, b, c) = (y, y, x)$ , we get $f(x, y) = f(y, x)$ for all $x, y$ . Consequently, if we define the function $g \colon \mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \to \mathbb{Z}_{\geq 0}$ , such that $g(a, b) := f(a, b) - 1$ for all non-negative integers $a$ and $b$ , then the original functional equation becomes

$g(a, b) + g(c, b) = g(ac, b^2).$
Let $P(a, b, c)$ denote the assumption on $(a, b, c)$ in this equation. For $b = 1$ , we get that the function $h(x) := g(x, 1)$ depends only on the value of $h(p)$ for all prime numbers $p$ . That is, by induction we have:

$h(x) = \sum_{p \text{ prime}} h(p)\nu_p(x). \tag{2}$
Now we will prove that $h(3) > 0$ or $h(5) > 0$ . Assume the contrary, i.e.

$h(45) = g(45, 1) = 2g(3, 1) + g(5, 1) = 0.$
Hence from $P(1, 45, 45)$ and $P(45, 1, 2025)$ follows that $g(45, 2025) = g(45, 45)$ . Also, from $P(1, 45, 2025)$ we get that $g(2025, 2025) = g(45, 2025) = g(45, 45)$ . But $g(2025, 2025) = 2g(45, 45)$ from $P(45, 45, 45)$ , from which $g(2025, 2025) = 0$ , contradiction. Therefore $h(3) > 0$ or $h(5) > 0$ .

The main substitution we use is $P(1, x, n)$ , which gives us $h(x) + g(n, x) = g(n, x^2)$ . Continuing with $P(1, x^2, n), P(1, x^4, n)$ , and so on, we inductively obtain:

$g(n, x) = g(n, x^n) + (x^2 - 1)h(x) + (x^4 - 1)h(x) + \ldots.$
Suppose $h(x) > h(x + 1)$ for some $x$ , then:

$g(n, x) + (x^2 - 1)h(x) = g(n, x^2) \le g(n, (x + 1)^2) = g(n, x + 1) + (x^2 - 1)h(x + 1).$
The latter cannot be true for all $k$ , because $h(x) > h(x + 1)$ , and $g(n, x), g(n, x + 1)$ are constants. Therefore $h(x) \le h(x + 1)$ for all $x \in \mathbb{N}$ from the beginning of the problem to case $n = 1$ . That is $h(3) \le h(5)$ or $h(5) \le h(3)$ , but $h(p) > 0$ for every prime $p \ge 5$ because $h$ is monotonically increasing. Let $h(5) = a, h(7) = b = 7h(5) > 1$ . Clearly, $a \ne b$ , so let us assume $a < b$ (the case $a > b$ is analogous). Then we have

$h(a^k) = kh(5^k)(7) = kh(5)(7) = h(b^k).$
Since $h$ is monotonically increasing, we obtain $h(a^k) = h(a^k + 1) = \ldots = h(b^k)$ . Let us choose a prime $q > h(5)(7)(k) = q + 1$ , such that

$q + 1 = \frac{a^{q + 1} - 1}{a - 1} > \frac{b^{q + 1} - 1}{b - 1} > b^q + 1.$
It follows that there exists a positive integer $a$ , such that $a^{q + 1} < b^{q + 1}$ . From above, we have $h(a^{q + 1}) = h(b^{q + 1}) = h(a^q) = h(b^q) = h(5^k)(7)$ , which contradicts the choice $q \ne q + 1$ and $h(1)(5)(7)$ . This completes the proof.

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