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

Home > Just Math
Always Divisible (Posted on 2007-05-20) Difficulty: 3 of 5
Prove that for every integer x, there is an integer y such that (y^2-2)/(x^4+1) is an integer.

See The Solution Submitted by Brian Smith    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Puzzle Solution Comment 3 of 3 |

At the outset, it is trivial to observe that for |x| =1, the quotient
(y^2-2)/2 is always an integer whenever y is even.

It is also trivial to observe that for x =0, the given quotient
reduces to (y^2-2) which is always an integer for any given integer t.

Thus, we will restrict ourselves to the case whenever |x|>=2

By the problem
(y^2-2)/(x^4+1) = p(say), where p is an integer.

Or, y^2 Mod (x^4+1) = 2

By inspection, we find that:

x=2 gives (y,p) = (6,2)
x=3 gives (y,p) = (24, 7)
x=4 gives (y,p) = (60, 14)

This leads us to conjecture that:
y Mod (x^4+1) = x^3-x
Or, y = (x^3-x) +t(x^4+1), where t is an integer......(**)

Substituting the value of y in relationship (**), we
observe that:

p = (y^2-2)/(x^4+1)
= (x^2-2) + 2t(x^3-x) + t^2(x^4+1)
>= (x^2-2) + (x^3-x)+(x^4+1), whenever t is non negative and x is positive
>= 2

The said quotient is thus always an integer for non negative t.

When t is negative, let us substitute t=-u,where u  is positive,
p = (x^2-2) - 2u(x^3-x) + u^2(x^4+1)
>= (x^4 -2*x^3+x^2+2x -1)
> 0, whenever x>= 2

Thus the case is true for x>2
In a similar vein, it can be shown that p is an integer whenever

Consequently, it follows that there always exist an integer y such that (y^2-2)/(x^4+1) is an integer.

Edited on May 21, 2007, 5:22 am
  Posted by K Sengupta on 2007-05-21 05:22:34

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