prove |a-b| is a square, given a-b=a^2c-b^2d

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

prove |a-b| is a square, given a-b=a^2c-b^2d

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Macau Inter High School Competition

The post below has been deleted. Click to close.

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

196 posts

#1 h Mar 7, 2020, 2:19 AM

Y by

Let $a, b$ be integers such that there are consecutive integers $c,d$ satisfy $a-b=a^2 c-b^2 d$ .
Prove : $|a-b|$ is a perfect square.

Z K Y

The post below has been deleted. Click to close.

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

MessingWithMath

340 posts

MessingWithMath

#2 h Mar 7, 2020, 2:56 AM • 1 Y

Y by Mango247

WLOG $a>b$ or we can just switch $c$ and $d$ around to make it work. Then, let $a-b=k>0$ as $0$ is already a perfect square. We see quickly that $d=c+1$ or else the $LHS\geq a^2-b^2 > a-b$ except for a few trivial cases.Then, solving for $c$ in terms of $k$ and $b$ , we have $c=\frac{k+b^2}{k(2b+k)}$ . Assume that a prime $p$ divides $k$ an odd number of times (call this power $m$ ). Then, since $b^2\equiv 0 \pmod{k}$ , we can assume $p^n|b$ . Thus, we have two cases, when $2n>m$ or $2n<m$ . If $2n>m$ , then $v_p(k+b^2)=m$ but $v_p(k(2b+k))>m$ so contradiction. Else, if $2n<m$ , then we have $v_p(k+b^2)=2n<m$ but $v_p(k(2b+k))>m$ so contradiction. Thus, we prove that this can't happen, so all prime factors of $k$ have even power. This means that $k$ is a perfect square.

Z K Y

The post below has been deleted. Click to close.

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

6107 posts

#3 h Mar 7, 2020, 3:43 AM

Y by

Alpha314159 wrote:

Let $a, b$ be integers such that there are consecutive integers $c,d$ satisfy $a-b=a^2 c-b^2 d$ .
Prove : $|a-b|$ is a perfect square.

This problem is also there in Mathematical Olympiad Challenges by Titu Andreescu

Z K Y

The post below has been deleted. Click to close.

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

196 posts

#4 h Mar 7, 2020, 10:05 AM

Y by

Thanks for the the help

Z K Y

The post below has been deleted. Click to close.

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

3 posts

#5 h Apr 17, 2025, 4:19 PM • 2 Y

Y by Omerking, aqusha_mlp12

Alpha314159 wrote:

Let $a, b$ be integers such that there are consecutive integers $c,d$ satisfy $a-b=a^2 c-b^2 d$ .
Prove : $|a-b|$ is a perfect square.

Let $gcd(a;b)=x$ then $a=xy$ , $b=xk$ where $gcd(y;k)=1$ $c=d+1$ So $y-k=xy^2d+xy^2-xk^2d$ .
We get $x|y-k$ .
$y-k=xd(y-k)(y+k)+xy^2$ $xy^2=(y-k)(1-xdy-xdk)$
$gcd(y;y-k)=1$
$gcd(1-dxy-dxk;x)=1$
We get $y-k= \pm x$
$|a-b|=x^2$ finished $d=c+1$ case is analogical

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a