Emphatic triples

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Emphatic triples

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: ELMO Shortlist 2024/N2

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4999 posts

#1 h Jun 22, 2024, 3:49 PM

Y by

Call a positive integer emphatic if it can be written in the form $a^2+b!$ , where $a$ and $b$ are positive integers. Prove that there are infinitely many positive integers $n$ such that $n$ , $n+1$ , and $n+2$ are all emphatic.

Allen Wang

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

834 posts

#2 h Jun 22, 2024, 4:04 PM • 1 Y

Y by MatSeFner

Consider $a=\frac{b!}{4}-1$ . Then, $a^2+b!$ , $(a+2)^2+1!$ , $(a+2)^2+2!$ works.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1349 posts

i3435

#3 h Jun 22, 2024, 4:09 PM • 2 Y

Y by LLL2019, Assassino9931

Not a solution but related: By $\mod 8$ and finite case check for small values, you can prove that at maximum $4$ consecutive numbers are emphatic. I don't know if there are infinitely $n$ with $n,n+1,n+2,n+3$ emphatic, but $n=168$ works.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1194 posts

#4 h Jun 22, 2024, 4:14 PM • 2 Y

Y by VicKmath7, Stuffybear

It suffices to show that there are infinitely many solutions to $a^2 + b! = m^2$ , as $1 = 1!$ and $2! = 2$ . We want $(m-a)(m+a) = b!$ , so take $b\geq 4$ and we can have $m-a = 2$ , $m+a = \frac{b!}{2}$ , i.e. $a = \frac{b!}{4} - 1$ , $m = \frac{b!}{4} + 1$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

406 posts

#5 h Jun 23, 2024, 11:55 AM

Y by

IAmTheHazard wrote:

Call a positive integer emphatic if it can be written in the form $a^2+b!$ , where $a$ and $b$ are positive integers. Prove that there are infinitely many positive integers $n$ such that $n$ , $n+1$ , and $n+2$ are all emphatic.

Allen Wang

$x^2+1=a^2+1!$ , $x^2+2=a^2+2!$
So we need to prove that there is infinite solution to the equation $x^2+3=a^2+b!$
For $x=a+c$ we get $c(c+2a)=b!-3$
And for $b>=3$ we can get $c=3$ and we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

5531 posts

#6 h Jul 11, 2024, 2:51 AM

Y by

Note that $a^2 + 1, a^2 + 2$ are clearly emphatic for any positive integer $a$ .

Claim: There are infinitely many perfect square that are emphatic.
Proof: For any $b > 2$ , we note that if $b! = 4k$ , then $(k+1)^2 = (k-1)^2 + b!$ , so $(k+1)^2$ is emphatic. There are clearly infinitely many possible values of $k$ , so the claim is proven. $\square$

Now, there exist infinitely many $a$ where $a^2$ is emphatic, and $a^2 + 1, a^2 + 2$ are also emphatic, so taking $n = a$ for any such $a$ works.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

76 posts

#7 h Aug 29, 2024, 3:57 PM

Y by

Let $d$ be a positive divisor of $\frac{m!}{4}$ where $m\geq 4$ is an integer. Let $n=(\frac{m!}{4d}+d)^2$ . Then the following equalities hold.

$n=(\frac{m!}{4d}-d)^2+m!$ $n+1=(\frac{m!}{4d}+d)^2+1!$ $n+2=(\frac{m!}{4d}+d)^2+2!$
Therefore $n$ is emphatic. Changing values of $m$ and $d$ gives us infinitely many emphatic positive integers. Therefore there exists infinitely many emphatic positive integers.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

58 posts

#8 h Oct 18, 2024, 3:55 PM

Y by

Very Easy.
Take $b$ such $2|b$ and $a=\frac{(b-1)!-b}{2}$
And take $n=a^2+b!$ .
Note that $n$ is a square number.
$n+1=n+1!$ and $n+2=n+2!$ so there exists infinite triplets $(n,n+1,n+2)$
$\square$

This post has been edited 1 time. Last edited by Warideeb, Oct 18, 2024, 3:56 PM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a