All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Perfect Square Not (Posted on 2008-05-07) Difficulty: 2 of 5
The pairs of positive integers (A, B) are such that B divides 2A2.

Prove that A2 + B cannot be a perfect square.

See The Solution Submitted by K Sengupta    
Rating: 2.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
To B not a square | Comment 2 of 6 |
Since we are told b divides 2a, then let k=2a/b, so k is an integer and kb=2a. It is to prove a+b is not a perfect square. To show this, it can be shown if it did, then it leads to a contradiction.

Suppose a+b=x for some integer x. Then subtract a from both sides and multiply by k to get kb=k(x-a)=k(x-a)(x+a). Substitute
kb=2a to get k(x-a)(x+a)=2a.

Now note g1=gcd(x-a,a) and g2=gcd(x+a,a) are equal, for positive integers x and a. To show this, note g1 divides x+a, since if g1 divides x-a and a, then it must divide x and thus x+a. Thus, g1 is a common divisor of x+a and a, so g1 divides g2. A similar proof can show g2 divides g1, thus g1=g2. Let g be this common gcd.

Then (x-a)/g and a are relatively prime, so ((x-a)/g) and a are relatively prime, and thus (x-a)/g and a are relatively prime. Similarly, (x+a)/g and a are relatively prime.

Next is to show that (x-a)/g and (x+a)/g are relatively prime to 2a. Let p be the largest power of 2 which divides a. Assume 2p divides (x-a), thus 2p must divide x+a=(x-a)+2a since 2p divides 2a. Thus, 2p*2p=4p divides 2a, but this contradicts p being the largest power of 2 in a. (It can be shown similarly that assuming 2p divides (x-a) leads to a contradiction.)

Thus, p is the largest power of 2 in x+a and x-a. This shows (x+a)/g and (x-a)/g must be odd since p divides g. Thus, (x-a)/g and (x+a)/g are relatively prime to 2a.

If (x-a)/g>1, or (x-a)/g>1 then x-a or x+a would contain a factor ((x-a)/g or (x+a)/g) that 2a would not. Since x>a  then (x-a)/g>=1 and (x+a)/g>=1, so (x-a)/g=(x+a)/g=1, and (x-a)/g=(x+a)/g implies x-a=x+a.

However, x-a=x+a implies -a=a, 2a=0, a=0, a contradiction since a and b must be positive. So if one has (a,b) such that b divides 2a, then a+b isn't a perfect square.

  Posted by Gamer on 2008-05-07 17:22:09
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information