Solve a^7(a-1)=19b(19b+2) over Z

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

Solve a^7(a-1)=19b(19b+2) over Z

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Turkey EGMO TST 2022 P3

The post below has been deleted. Click to close.

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

577 posts

#1 h Mar 16, 2022, 1:05 PM • 1 Y

Y by teomihai

Find all pairs of integers $(a,b)$ satisfying the equation $a^7(a-1)=19b(19b+2)$ .

Z K Y

The post below has been deleted. Click to close.

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

577 posts

#2 h Mar 16, 2022, 1:06 PM • 1 Y

Y by ladyMS

We have
$(19b+1)^2=a^8-a^7+1=(a^2-a+1)(a^6-a^4-a^3+a+1)$ Also $gcd(a^2-a+1,a^6-a^4-a^3+a+1)=1,\; 19$ . But, $19$ doesn’t divide $(19b+1)^2$ . Hence, the gcd should be $1$ . This means both of the numbers are perfect squares.
Then, $(2a-1)^2+3$ is a perfect square. Hence, $a=0,1$ and for these values $b=1$ .

Z K Y

The post below has been deleted. Click to close.

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

Mahdi_Mashayekhi

694 posts

Mahdi_Mashayekhi

#3 h Apr 1, 2022, 2:59 PM • 1 Y

Y by Ismayil_Orucov

$x(x+2)$ reminds of $(x+1)^2 - 1$ so we have $a^8 - a^7 + 1 = (19b+1)^2$ . from here we need to rewrite $a^8 - a^7 + 1 = xy$ such that $\gcd{(x,y)} = 1$ . we know $a^8 - a^7 + 1 = (a^2-a+1)(a^6 - a^4 - a^3 + a + 1)$ and also with Euclidean algorithm we have $\gcd{(a^2-a+1,a^6 - a^4 - a^3 + a + 1)} = {1,19}$ . Note that $(19b+1)^2 = 1$ mod $(19)$ so $\gcd{(a^2-a+1,a^6 - a^4 - a^3 + a + 1)} = 1$ so both $a^2-a+1,a^6 - a^4 - a^3 + a + 1$ are squares. we have $(a-1)^2 \le a^2 - a + 1 \le a^2$ so only cases which $a^2 - a + 1$ is square is when equalities hold so $a = {0,1}$ , either one we have $(19b+1)^2 = 1 \implies b = 0$ .
Answers : $(a,b) = (0,0)$ .

Z K Y

The post below has been deleted. Click to close.

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

14 posts

#4 h Apr 4, 2025, 2:42 PM

Y by

$a^7(a-1)=19b(19b+2) \implies a^7(a-1)+1=(19b+1)^2$ .
So we can see $(19b+1)^2=a^8-a^7+1=(a^2-a+1)(a^6-a^4-a^3+a+1$ and $gcd(a^2-a+1,a^6-a^4-a^3+a+1)=1,19$ but $gcd(a^2-a+1,a^6-a^4-a^3+a+1)=1$ because $(19b+1)^2 \equiv 0(mod 19)$ . I mean $a^2-a+1$ and $a^6-a^4-a^3+a+1$ are perfect squares. $a^2 \le a^2-a+1 \le (a+1)^2$ . a should be 0 or 1 because of $a^2 \le a^2-a+1 \le (a+1)^2$ . We have two solution. These are $(a,b)=(0,0),(1,0)

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a