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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Factorial Divisibility
Aryan-23   47
N a minute ago by ezpotd
Source: IMO SL 2022 N2
Find all positive integers $n>2$ such that
$$ n! \mid \prod_{ p<q\le n, p,q \, \text{primes}} (p+q)$$
47 replies
Aryan-23
Jul 9, 2023
ezpotd
a minute ago
2-var inequality
sqing   3
N 25 minutes ago by sqing
Source: Own
Let $ a,b> 0 ,a^3+ab+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq 8$$$$ (a^2+b^2)(a+1)(b+1) \leq 8$$Let $ a,b> 0 ,a^3+ab(a+b)+b^3=3.$ Prove that
$$ (a+b)(a+1)(b+1) \leq \frac{3}{2}+\sqrt[3]{6}+\sqrt[3]{36}$$
3 replies
1 viewing
sqing
an hour ago
sqing
25 minutes ago
Infinite number of sets with an intersection property
Drytime   8
N 42 minutes ago by math90
Source: Romania TST 2013 Test 2 Problem 4
Let $k$ be a positive integer larger than $1$. Build an infinite set $\mathcal{A}$ of subsets of $\mathbb{N}$ having the following properties:

(a) any $k$ distinct sets of $\mathcal{A}$ have exactly one common element;
(b) any $k+1$ distinct sets of $\mathcal{A}$ have void intersection.
8 replies
Drytime
Apr 26, 2013
math90
42 minutes ago
Factorials divide
va2010   37
N an hour ago by ND_
Source: 2015 ISL N2
Let $a$ and $b$ be positive integers such that $a! + b!$ divides $a!b!$. Prove that $3a \ge 2b + 2$.
37 replies
va2010
Jul 7, 2016
ND_
an hour ago
IMC 1994 D2 P1
j___d   13
N Yesterday at 11:20 PM by krigger
Let $f\in C^1[a,b]$, $f(a)=0$ and suppose that $\lambda\in\mathbb R$, $\lambda >0$, is such that
$$|f'(x)|\leq \lambda |f(x)|$$for all $x\in [a,b]$. Is it true that $f(x)=0$ for all $x\in [a,b]$?
13 replies
j___d
Mar 6, 2017
krigger
Yesterday at 11:20 PM
D1039 : A strange and general result on series
Dattier   0
Yesterday at 10:33 PM
Source: les dattes à Dattier
Let $f \in C([0,1];[0,1])$ bijective, $f(0)=0$ and $(a_k) \in [0,1]^\mathbb N$ with $ \sum \limits_{k=0}^{+\infty} a_k$ converge.

Is it true that $\sum \limits_{k=0}^{+\infty} f(a_k)\times f^{-1}(a_k)$ converge?
0 replies
Dattier
Yesterday at 10:33 PM
0 replies
Aproximate ln(2) using perfect numbers
YLG_123   5
N Yesterday at 8:55 PM by ei_killua_
Source: Brazilian Mathematical Olympiad 2024, Level U, Problem 1
A positive integer \(n\) is called perfect if the sum of its positive divisors \(\sigma(n)\) is twice \(n\), that is, \(\sigma(n) = 2n\). For example, \(6\) is a perfect number since the sum of its positive divisors is \(1 + 2 + 3 + 6 = 12\), which is twice \(6\). Prove that if \(n\) is a positive perfect integer, then:
\[
\sum_{p|n} \frac{1}{p + 1} < \ln 2 < \sum_{p|n} \frac{1}{p - 1}
\]where the sums are taken over all prime divisors \(p\) of \(n\).
5 replies
YLG_123
Oct 12, 2024
ei_killua_
Yesterday at 8:55 PM
Quadruple Binomial Coefficient Sum
P162008   4
N Yesterday at 8:40 PM by vmene
Source: Self made by my Elder brother
$\sum_{p=0}^{\infty} \sum_{r=0}^{\infty} \sum_{q=1}^{\infty} \sum_{s=0}^{p+q - 1} \frac{((-1)^{p+r+s+1})(2^{p+q-1}) \binom{p + q - s - 1}{p + q - 2s - 1}}{4^s(2p^2q + 2pqr + pq + qr)(2p + 2q + 2r + 3)}.$
4 replies
P162008
Thursday at 8:04 PM
vmene
Yesterday at 8:40 PM
IMC 1994 D1 P5
j___d   5
N Yesterday at 5:39 PM by krigger
a) Let $f\in C[0,b]$, $g\in C(\mathbb R)$ and let $g$ be periodic with period $b$. Prove that $\int_0^b f(x) g(nx)\,\mathrm dx$ has a limit as $n\to\infty$ and
$$\lim_{n\to\infty}\int_0^b f(x)g(nx)\,\mathrm dx=\frac 1b \int_0^b f(x)\,\mathrm dx\cdot\int_0^b g(x)\,\mathrm dx$$
b) Find
$$\lim_{n\to\infty}\int_0^\pi \frac{\sin x}{1+3\cos^2nx}\,\mathrm dx$$
5 replies
j___d
Mar 6, 2017
krigger
Yesterday at 5:39 PM
2023 Putnam A2
giginori   21
N Yesterday at 3:32 PM by pie854
Let $n$ be an even positive integer. Let $p$ be a monic, real polynomial of degree $2 n$; that is to say, $p(x)=$ $x^{2 n}+a_{2 n-1} x^{2 n-1}+\cdots+a_1 x+a_0$ for some real coefficients $a_0, \ldots, a_{2 n-1}$. Suppose that $p(1 / k)=k^2$ for all integers $k$ such that $1 \leq|k| \leq n$. Find all other real numbers $x$ for which $p(1 / x)=x^2$.
21 replies
giginori
Dec 3, 2023
pie854
Yesterday at 3:32 PM
Putnam 2019 A1
awesomemathlete   33
N Yesterday at 3:25 PM by cursed_tangent1434
Source: 2019 William Lowell Putnam Competition
Determine all possible values of $A^3+B^3+C^3-3ABC$ where $A$, $B$, and $C$ are nonnegative integers.
33 replies
awesomemathlete
Dec 10, 2019
cursed_tangent1434
Yesterday at 3:25 PM
IMC 1994 D1 P2
j___d   5
N Yesterday at 3:11 PM by krigger
Let $f\in C^1(a,b)$, $\lim_{x\to a^+}f(x)=\infty$, $\lim_{x\to b^-}f(x)=-\infty$ and $f'(x)+f^2(x)\geq -1$ for $x\in (a,b)$. Prove that $b-a\geq\pi$ and give an example where $b-a=\pi$.
5 replies
j___d
Mar 6, 2017
krigger
Yesterday at 3:11 PM
A Construction in Multivariable Analysis
MrOrange   0
Yesterday at 2:11 PM
Source: Garling's A COURSE IN MATHEMATICAL ANALYSIS
Construct a continuous real valued function \( f \) on \( \mathbb{R}^2 \) for which
\[
\lim_{R \to \infty} \int_{\|x\|_2 \leq R} f(x) \, dx = 0
\]and for which
\[
\lim_{R \to \infty} \int_{\|x\|_\infty \leq R} f(x) \, dx \text{ does not exist.}
\]
0 replies
MrOrange
Yesterday at 2:11 PM
0 replies
Possible values of determinant of 0-1 matrices
mathematics2004   4
N Yesterday at 1:56 PM by loup blanc
Source: 2021 Simon Marais, A3
Let $\mathcal{M}$ be the set of all $2021 \times 2021$ matrices with at most two entries in each row equal to $1$ and all other entries equal to $0$.
Determine the size of the set $\{ \det A : A \in M \}$.
Here $\det A$ denotes the determinant of the matrix $A$.
4 replies
mathematics2004
Nov 2, 2021
loup blanc
Yesterday at 1:56 PM
Iran second round 2025-q1
mohsen   8
N May 18, 2025 by Autistic_Turk
Find all positive integers n>2 such that sum of n and any of its prime divisors is a perfect square.
8 replies
mohsen
Apr 19, 2025
Autistic_Turk
May 18, 2025
Iran second round 2025-q1
G H J
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mohsen
16 posts
#1 • 1 Y
Y by sami1618
Find all positive integers n>2 such that sum of n and any of its prime divisors is a perfect square.
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sami1618
920 posts
#2 • 1 Y
Y by Sedro
Answer: No such number exists.

Solution. We start by proving the following two claims:

Claim 1. We have that $n$ is squarefree.
Proof. If there exists a prime $p$ for which $p^2|n$, then $n+p$ must be a perfect square. But $v_p(n+p)=1$, a contradiction.

Claim 2. We have that $n$ has at most two distinct prime divisors.
Proof. Assume otherwise that $p$, $q$, and $r$ are three distinct prime divisors of $n$. Without loss of generality, assume that $p<q<r$. It is easy to see that both $p$ and $q$ must be strictly smaller than $\sqrt{n}$. On the other hand, $$\sqrt{n+q}-\sqrt{n+p}\geq 1\Rightarrow q-p\geq \sqrt{n+q}+\sqrt{n+p}\geq 2\sqrt{n},$$which is a contradiction.

By the Claims, we must have that either $n=p$ for some prime $p>2$ or $n=pq$ for some distinct primes $p$ and $q$. In the first case, $2p$ must be a perfect square, which is absurd. In the second case, both $p(q+1)$ and $q(p+1)$ must be perfect squares. This implies that $p|q+1$ and that $q|p+1$. Assume that $p<q$. Then it must be that $p=q-1$ which implies that $p=2$ and $q=3$, which fails to work.
This post has been edited 1 time. Last edited by sami1618, Apr 19, 2025, 5:33 PM
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missionjoshi.65
2 posts
#3 • 2 Y
Y by Calc_1, sami1618
Posting though very similar as above!
Claim 1: $n$ can't be a power of prime.

Suppose, $n = p^k$ for some prime $p$, then we have $v_p(p^k+p) =1+v_p(p^{k-1}+1),$ which can't be a square when $k \geq 2$. And, $k = 1$ gives $n = 2$ which is a contradiction.

Claim 2: $n$ cannot have three or more prime factors.

Suppose $p,q,r | n$ and $p<q<r$, then

$$n +p = a^2 \quad \text{and} \quad n+q = b^2$$
This implies:
$$ q-p = (b-a)(b+a)$$
Similarly,
$$ q-p \geq b+a = \sqrt{n+p} + \sqrt{n+q} > \sqrt{n}+\sqrt{n} = 2 \sqrt{n}$$
which is a contradiction since both $p, q < \sqrt{n}.$

Now, for two primes $p<q$ we have $p|q+1$ and $q|p+1$ which implies $p=q-1,$ so only possible values are $p=2$ and $q=3$, which do not work. And, we are done!
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Parsia--
79 posts
#4 • 2 Y
Y by Mahdi_Mashayekhi, amirhsz
Here's a much different and overkill approach.
suppose $n$ has at least two different primes $q,r$ such that $q \equiv r \equiv 3 \mod 4$. Note that $n$ is also square-free.
So we get
$$q(\frac{n}{q}+1) = x^2 \Rightarrow (\frac{r}{q})=1 $$Similarly $ (\frac{q}{r})=1$.
Now note that $$ (\frac{r}{q})(\frac{q}{r})=(-1)^{\frac{r-1}{2}. \frac{q-1}{2}}=-1 \Rightarrow  1=(\frac{r}{q})=-(\frac{q}{r})=-1$$which is a contradiction.
if $n = 2p$, we must have $3p$ is a perfect square which gives $p=3$ which doesn't work.
Now $n$ odd won't work because of mod $4$. so $n$ is even. Consider a prime divisor $p$ of $n$ such $p \equiv 1 \mod 4$ we get $p(\frac{n}{p}+1) =x^2$ but $\frac{n}{p}+1 \equiv 3 \mod 4$.
Thus we conclude that no such $n$ exists.
This post has been edited 2 times. Last edited by Parsia--, May 10, 2025, 3:18 PM
Reason: Shortened the solution
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MathLuis
1557 posts
#5
Y by
Suppose FTSOC there existed such $n$, then if $n$ was not squarefree then there exists a prime $p$ with $p^2 \mid n$ however we have $n+p$ is a perfect square and thus $2 \mid \nu_p(n+p)=1$, a clear contradiction!, therefore $n$ is squarefree.
Now let $n=p_1 \cdot p_2 \cdots p_k$ where $p_1<p_2 \cdots p_k$. Clearly $k \ge 2$ as if $k=1$ then $2p_1$ is a perfect square but $p_1 \ne 2$ in this case so this isn't true, and if $k \ge 3$ then notice we must have $\sqrt{n+p_2}-\sqrt{n+p_1} \ge 1$ which implies that $n+p_2 \ge n+p_1+2\sqrt{n+p_1}+1$ which implies that $p_2^2>(p_2-p_1-1)^2 \ge 4n+4p_1>4p_2^2+4p_1$ which is clearly a contradiction.
And now if $n=p_1p_2$ then $p_2(p_1+1), p_1(p_2+1)$ are both perfect squares then $p_2 \mid p_1+1$ and $p_1 \mid p_2+1$ so $p_1p_2 \mid p_1p_2+p_1+p_2+1$ which means $p_1p_2 \mid p_1+p_2+1$ which means that $2 \le (p_1-1)(p_2-1) \le 2$ where equality is attained when $p_1=2$ and $p_2=3$ however $2 \cdot 4=8$ is not a perfect square thus contradiction either way and we are done :cool:
This post has been edited 1 time. Last edited by MathLuis, Apr 23, 2025, 12:31 AM
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math-helli
13 posts
#6
Y by
Here you can find some solutions of second round
https://t.me/matholampiad123
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Marco22
4 posts
#7
Y by
Answer:such number do not exist.
Suppose on the contrary that a number $n$ satisfying the properties exists, and let $p$ and $q$ be two prime divisors of $n$. Then by the hypothesis there exist $x,y \in \mathbb{N}$ such that $n+p=x^2$ and $n+q=y^2$. clearly we can see that $p|x$ and $p|y$. Then by a size argument it follows:
$p-q=x^2-y^2=(x-y)(x+y)=(ap-bq)(ap+bq)>p+q$ for some integers $a$ and $b$. So the answer clearly follows
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Mathgloggers
92 posts
#8
Y by
Notice that each prime sums with $n$ to a distinct perfect square.
So let $k \in N$ such that $k^2>n$
We have, $(k+3)^2-n<\frac{n}{2} $,because the max prime is $< \frac{n}{2}$
$k^2+9+6k<\frac{3n}{2} $ $\implies$ $n^2<k<\frac{\frac{n}{2}-9}{6} $,now just solve this equation for $n$ you will get $-2<n<2$ ,So you it also does not work.

So it has clearly at most 2 prime factors for $n=p.q>1$ which respectively sums to $n+p=k^2$ and $n+q=(k+1)^2$.
hence one has to be even =$2$ and the other must be $2k+3$ and $n=k^2-2$



Now just put all in that equation you will get :$k^2-4k-8=0$ no integer solution .
(I don't why my solution was so basic:(
This post has been edited 1 time. Last edited by Mathgloggers, May 18, 2025, 6:41 AM
Reason: m
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Autistic_Turk
12 posts
#9
Y by
mohsen wrote:
Find all positive integers n>2 such that sum of n and any of its prime divisors is a perfect square.

Here is my solution that I actually used in contest
Trivialy there exist positive intiger m and non negative integer q so that n=m^2+q and n<(m+1)^2 so q<2m+1 now there are 2 cases
Case 1: n has at least two different prime divisor
Let p_1 and p_2 be prime divisor of n and WLOG p_2>p_1 notice there exist intiger r_1 and r_2 such that n+p_1=(m+r_1)^2 and n+p_2=(m+r_2)^2 and p_2 divide n therfore p_2 divide n+p_2 so p_2 divide (m+r_2)^2 thus p_2 divide m+r_2 but we know p_2=(m+r_2)^2-n=m^2+2(r_2)m+(r_2)^2-m^2-q and we know any natural divisor of a natural number is smaller or equal to that number so we have
2(r_2)m +(r_2)^2-q =< m+r_2
But we have r_2=>2 and q<=2m+1 thus contradiction is trivial
Case 2: n has only one prime factor
Thus there exist prime number p and positive intiger a such that n=p^a therfore p^a+p is square but for any a larger then 1 we know that p divide n and p^2 do not divide n thus n can't be a square so a=1 and for a=1 we would get 2p is square thus 4 divide 2p and p is even thus p=2
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