Find a solution to:
x
1^4 + x
2^4 + x
3^4 + ... + x
n^4 = 1999
where each xy is a distinct integer.
(Or prove that it is impossible).
(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.
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Posted by Richard
on 2004-09-12 18:49:04 |
re(2): Solution
