Stereotypical Diophantine Equation

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

Stereotypical Diophantine Equation

G H J

Reveal topic tags

Diophantine equation

L

G H BBookmark kLocked kLocked NReply

Source: 2025 Nepal Mock TST Day 2 Problem 1

The post below has been deleted. Click to close.

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

1469 posts

#1 h Apr 6, 2025, 1:32 PM • 1 Y

Y by khan.academy

Find all solutions in the nonnegative integers to $2^a3^b5^c7^d - 1 = 11^e$ .

(Shining Sun, USA)

Z K Y

The post below has been deleted. Click to close.

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

69 posts

kokcio

#2 h Apr 6, 2025, 1:42 PM

Y by

Firstly, we can see that $11$ is quadratic residue modulo $7$ , but $-1$ isn't, so we cannot have that $7|11^e+1$ .
We also have that $11^e+1\equiv 2 \mod 5$ , so we cannot have $5|11^e+1$ .
From this two observations, we have that $c=d=0$ . Now from Zsigmondy theorem we know that if $e>1$ then there exist prime $p>3$ such that $p|11^e+1$ , so we have to have $e\leq 1$ . Therefore, we find two solutions $(a,b,c,d,e)=(1,0,0,0,0)$ and $(a,b,c,d,e)=(2,1,0,0,1)$

Z K Y

The post below has been deleted. Click to close.

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

1413 posts

#3 h Apr 6, 2025, 1:57 PM

Y by

Here's a solution without Zsigmondy.

Notice first that if $c\ge 1$ then we have a contradiction via modulo 5, whereas if $d\ge 1$ then we have a contradiction via modulo 7. So, $c=d=0$ and $2^a3^b-1=11^e$ . Next, suppose $b=0$ and $2^a=11^e+1$ . If $e$ is odd, then RHS is divisible by 3, a contradiction. For $e$ even, $11^e+1\equiv 2\pmod{4}$ , so $a=1$ . This gives the pentuple $(a,b,c,d,e)=(1,0,0,0,0)$ . In the remainder, assume $b\ge 1$ . Using modulo 3, we find $e$ is odd. This, in turn, implies $11^e+1\equiv 4\pmod{8}$ , forcing $a=2$ :
$4\cdot 3^b-1 = 11^e.$ For $b=1$ , we get the pentuple $(a,b,c,d,e)=(2,1,0,0,1)$ . Assume $b\ge 2$ . Using modulo 9, we obtain that $e\equiv 3\pmod{6}$ . Set $e=3u$ for $u$ odd. We obtain $4\cdot 3^b-1 = (11^u+1)(11^{2u}-11^u+1)$ . Check that $11^{2u}-11^u+1$ is odd, so it must be a power of two. However, for any $x$ , $x^2-x+1$ is not divisible by $9$ (easy to see), so $11^{2u}-11^u+1\in\{1,3\}$ , both of which are impossible. This completes the proof.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a