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

Home > Numbers
Fermat for beginners (Posted on 2006-09-14) Difficulty: 3 of 5
Determine all positive integer solutions of x2y+(x+1)2y=(x+2)2y

Of course, for y>1 you could use Fermat's theorem, but that would be unsporting!

No Solution Yet Submitted by JLo    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution, without invoking Fermat Comment 4 of 4 |
This puzzle was pretty easy after I found the right method.  Finding that method took far longer than I thought it would.  The key is the identity a^n - b^n = [a - b] * [a^(n-1) + a^(n-2)*b + a^(n-3)*b^2 + ... + b^(n-1)].

Rearrange the equation as (x+2)^(2y) - (x+1)^(2y) = x^(2y)

Factor the left side as [(x+2)^2 - (x+1)^2] * [(x+2)^(2y-2) + (x+2)^(2y-4)(x+1)^2 + (x+2)^(2y-6)(x+1)^4 + ... + (x+1)^(2y-2)].
This corresponds to a=(x+2)^2 and b=(x+1)^2 in the identity given at the beginning.

Then (x+2)^2 - (x+1)^2 = 2x + 3 must be a factor of x^(2y).

2x+3=x^(2y) mod x implies 3=0 mod x.  Then x is a factor of 3, namely 1 or 3.

Checking x=1: 1^(2y) + 2^(2y) = 3^(2y).  This works if y=1/2, but y must be a positive integer.

Checking x=3: 3^(2y) + 4^(2y) = 5^(2y).  This works for y=1.

The only solution is x=3, y=1 forming the equation 3^2 + 4^2 = 5^2.

  Posted by Brian Smith on 2016-06-26 10:34:55
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 (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

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