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Gonna party like it's 1999 (Posted on 2004-09-12) Difficulty: 1 of 5
Find a solution to:
x1^4 + x2^4 + x3^4 + ... + xn^4 = 1999

where each xy is a distinct integer.

(Or prove that it is impossible).

See The Solution Submitted by SilverKnight    
Rating: 3.2500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Solution | Comment 6 of 8 |
(In reply to re: Solution by David Shin)

You seem to have confused the x_i with their 4th powers in your writeup, but I see what you mean. A slightly different viewpoint, but using your mod 16 idea, is to write the original equation as

     summation from n=-6 to 6 of b_n*n^4 = 1999,

where the b_n are 0 or 1. Taking this modulo 16 then gives us

     b_(-5)+b_(-3)+b_(-1)+b_1+b_3+b_5 = 15 mod 16

so some number between 0 and 6 would then have to be congruent to 15 mod 16, a clear contradiction.


  Posted by Richard on 2004-09-12 18:49:04
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