Track / Stage 3 / 123 of 260 #603 of 2444

Problem 603

AMC 10/12, early questions

Number theory Difficulty 3.7 Prove it Junior Macedonian Mathematical Olympiad · North Macedonia

Solve the equation
$x_1^4 + x_2^4 + \dots + x_{14}^4 = 2016^3 - 1.$
in the set of integers.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

For $x = 2k$ , $x^4 = 16k^4 \equiv 0 \pmod{16}$ . For $x = 2k+1$ , $x^4 - 1 = 8k(k+1)(2k^2 + 2k + 1) \equiv 0 \pmod{16}$ , i.e. $x^4 \equiv 1 \pmod{16}$ . Since $2016^3 - 1 \equiv 15 \pmod{16}$ , and the sum of the numbers on the left-hand side never gives a remainder $15$ when divided by $16$ , it follows that the given equation has no solution in the integers.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.