About the problem
This problem asks for a counterexample to Euler’s sum of powers conjecture. Specifically, the problem asks: for some \(k > 5\), find nonzero integers \(a_1, \ldots, a_k\) such that \(\sum_{i=1}^{k-1} a_i^k = a_k^k\).
Euler conjectured that no solution to the above equation exists for \(k > 2\). The conjecture is true for \(k = 3\) (which is a special case of Fermat’s Last Theorem), but false for \(k = 4\) and \(k=5\). No counterexample for \(k > 5\) is known.
Experts expect that this problem is solvable, and the verifier can check a correct counterexample. The main risk with this problem is a brute-force solution for a single value of \(k\) may not be very interesting or insightful. An ideal solution would either be a construction that works for all \(k > 5\), or an efficient way of generating counterexamples.
Prompt
In 1769 Leonhard Euler conjectured that for every integer n > 2, no sum of fewer than n positive n-th powers can itself be an n-th power. In its tightest form this says the equation
a_1^n + a_2^n + ... + a_{n-1}^n = b^n
has no solution in positive integers: with only n-1 terms on the left, an n-th power cannot be reached. Euler's conjecture stood for nearly two centuries. It was finally refuted for n = 5 by Lander and Parkin in 1966, who found
27^5 + 84^5 + 110^5 + 133^5 = 144^5,
and later for n = 4 by Elkies and Frye. But for every exponent n > 5, no counterexample is known, and whether any exists remains open.
## Task
Find an integer k > 5 and nonzero integers a_1, ..., a_{k-1}, a_k such that
a_1^k + a_2^k + ... + a_{k-1}^k = a_k^k.
The length of your list determines the exponent: if you submit k integers, the exponent is k, the first k-1 are summed, and the last is the right-hand side. So a valid submission of k integers is exactly a counterexample to Euler's conjecture for exponent k.
## Submission
Submit your list of integers (a_1, ..., a_k) using the submit tool. The integers must be nonzero but may be negative. There is no bound on their magnitude. Your submission is checked exactly with arbitrary-precision integer arithmetic, so an approximate or rounded answer will not be accepted.