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

Home > Numbers
Further Arithmetic Integers (Posted on 2006-06-27) Difficulty: 3 of 5
Consider three positive integers x < y < z in arithmetic sequence, and determine all possible solutions of: x4 + y4 = z4 - 64

  Submitted by K Sengupta    
Rating: 2.5000 (2 votes)
Solution: (Hide)
(x,y,z) = (1, 2, 3) and (7, 8, 9) constitutes the only possible solution to the given equation.

EXPLANATION:

By the conditions of the problem:
z = y + a and x = y ľa , where a is any positive integer less than x.

Accordingly, in terms of provisions of the problem under reference:

(y + a)^4 - 64 = (y - a)^ 4 + y^4
or, 8*(y^3)* a + 8*(a^3)*y = y^4 + 64-------(#)
It is clearly observed from (#) that y is even ,so that substituting y = 2*p , we obtain : 4*(p^3)*a + (a^3)*p - 4 = p^4
Or, p(4*(p^2)*a + a^3 - p^3) = 4; so that p must be a positive divisor of 4 giving p= 1,2,4

Now; p=1 gives, a^3 + 4*a - 5 = 0, which yields:
(a-1)((a^2)+ a + 5) =0; giving a = 1, (-1 +/- sqrt(-19))/2, so that the only integral root in this case is given by a = 1
Hence, y=2*p = 2
Consequently, z= y+1 = 3 and x = y -1 = 1

p=2 gives, (a^3) + 16*a - 10 = 0, which does not possess any positive integer solution in a.

Now; p=4 gives, 4*(a^3) + 256*a - 4 = 256
or, (a^3) + 64*a - 65 = 0, giving:
(a-1)*((a^2) +a + 65)= 0, giving a=1 or, a= (-1+/- sqrt(-16899))/2.
Since, a is a positive integer, it follows that a=1; giving:
y=2*p = 2*4 = 8.

Consequently, z= y+1 = 9 and x = y -1 = 7

Therefore, (x, y, z) =(1, 2, 3) and (7, 8, 9) constitutes the only positive integral solutions to the equation under reference.

---------------------------------------------------------

Also refer to Bractals' solution and explanation.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
genearl problem :-DDaniel2006-06-28 22:53:13
re(2): SolutionBractals2006-06-28 10:53:32
re(2): SolutionBractals2006-06-28 10:32:38
Questionre: SolutionYosippavar2006-06-28 07:21:43
SolutionNot a Logical answerYosippavar2006-06-28 07:14:29
mathematica solution (spoiler)Daniel2006-06-28 06:03:32
Questionre: SolutionTrevor2006-06-27 21:37:31
SolutionSolutionBractals2006-06-27 21:01:09
re: not a proof but may be spoilerTrevor2006-06-27 16:16:56
Hints/Tipsnot a proof but may be spoilerCharlie2006-06-27 16:02:16
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information