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).
At the outset, we note that:
(2n)^4 = 16 * n^4 ==0 (Mod 16), for any integer n
Also, (2n+1)^2=4n^2+4n+1
= 8M+1, where M=n(n+1)/2, and n(n+1) must be even, being the product of consecutive integers.
So,
(2n+1)^4 =(8M+1)^2
= 64 *M^2 +16* M +1
= 16M(4M+1) +1
Hence, (2n+1)^4 == 1 (Mod 16)
Accordingly, the quartic residue of x(i)^4 is 0 or 1 accordingly as x(i) is even or odd for i=1 to n. .....(#)
Let n=14
Then, the maximum quartic residue of (#) of Sum x(i)^4 is 14.
i=1 to 14
Accordingly, there exists precisely 7 odd integers the maximum of which is 7
However, we note that:
7^4=2401> 1999
This is a contradiction.
So, no solution exists for n=14
For n> 14, the maximum odd integer must exceed 7.
Consequently, there does NOT exist any integer solution to the given problem.
NOTE
I am not happy with the solution. The general case of inadmissibility of non-distinct solution has only been proved here for n=14.
For general n, this proof is only true where i's are all distinct.
So, I have to look for non-distinct solutions for general n.
Edited on June 20, 2022, 10:02 pm
Puzzle Solution
