Competition problem — Number theory

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Number theory Difficulty 4.4 AIME Prove it United States

Problem:

Are there integers $a, b, c, d$ which satisfy $a^{4}+b^{4}+c^{4}+2016=10 d$ ?

Solution

Solution:

The answer is no. Look at the equation in base $5$ . Observe that $0^{4}=0$ , $1^{4}=1=1_{5}$ , $2^{4}=16=31_{5}$ , $3^{4}=81=311_{5}$ , $4^{4}=256=2011_{5}$ , so each of $a^{4}, b^{4}, c^{4}$ must end in $0$ or $1$ in base $5$ . On the other hand $10 d - 2016$ ends with $4$ in base $5$ . This is impossible.

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