Equation with powers

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

Equation with powers

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Serbia JBMO TST 2024 P1

The post below has been deleted. Click to close.

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

678 posts

#1 h May 25, 2024, 5:27 PM

Y by

Find all non-negative integers $x, y$ and primes $p$ such that $3^x+p^2=7 \cdot 2^y.$

Z K Y

The post below has been deleted. Click to close.

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

1133 posts

#2 h May 25, 2024, 6:10 PM

Y by

a_507_bc wrote:

Find all non-negative integers $x, y$ and primes $p$ such that $3^x+p^2=7 \cdot 2^y.$

If $y=0$ then $x=1, p=2$ . If $y=1$ then by mod 3 $x=0$ and there are no solutions. Now let $y>1$ , so $4 | RHS$ .
Note that since $4|7-3$ we have $2 \not | x$ . After this look at mod 8 to get $y<3$ ( $p \not = 2$ ).

This post has been edited 2 times. Last edited by NO_SQUARES, May 25, 2024, 7:30 PM
Reason: was wrong

Z K Y

The post below has been deleted. Click to close.

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

4918 posts

#3 h May 25, 2024, 7:21 PM • 2 Y

Y by NO_SQUARES, ehuseyinyigit

$p=2 \to y=0,x=1$
$p$ is odd $\to 3^x+p^2 \equiv 2,4 \pmod {8} \to y<3$
for $y=1$ there are no solutions
For $y=2: x=1,p=5$

Z K Y

The post below has been deleted. Click to close.

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

1364 posts

#4 h May 25, 2024, 8:20 PM

Y by

We only need that $p$ is odd if $p \geq 3$ . Indeed, mod 8 we have $3^x \equiv 3,1$ and $p^2\equiv 1$ and so $3^x + p^2 \equiv 2, 4$ while $7 \cdot 2^y \equiv 0$ for $y\geq 3$ , contradiction. Hence either $p=2$ or $y\leq 2$ .

If $p=2$ , then parity insists on $y=0$ , so $x=1$ . If $y=2$ , then only $p=5$ and $x=1$ works. If $y=1$ , then there are no solutions. If $y=0$ , then $p=2$ and $x=1$ works.

Hence all solutions are $(x,y,p) = (1,0,2), (1,2,5)$ .

Z K Y

The post below has been deleted. Click to close.

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

1 post

#5 h Jul 31, 2024, 4:16 PM • 1 Y

Y by JelaByteEngineer

Simply by applying mod 8 which restricts or bounds the value of y i.e y<3
By checking manually which gives us 2 solutions i.e (1,2,0);(1,2,5)

Z K Y

The post below has been deleted. Click to close.

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

52 posts

#6 h Mar 18, 2025, 4:34 PM

Y by

sketch of my solution

Z K Y

The post below has been deleted. Click to close.

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

71 posts

EVKV

#7 h Apr 3, 2025, 2:32 AM

Y by

p is odd for $y \neq 0$
and clearly x= 1 , y= 0, p=2 satisfies
$For y \geq 3$
$either 2,4 \equiv 0$ $mod$ $8$ nonsense
now checking remaining
all solutions are $(x,y,p) = (1,0,2), (1,2,5)$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a