WMC 2018 Senior Problem 18

By Euler's Totient Theorem, $a^{\varphi(8)}=a^4\equiv 1\pmod{8}$ since $a$ and $8$ are coprime. So $a^8\equiv 1\pmod{8}$ . Repeating this process for $b$ and $c$ , we find that $a^8+b^8+c^8\equiv 1+1+1=3\pmod{8}$ .

We note that $\,(2k+1)^2 = 4k(k+1)+1$ where is $k\geq 0$ is any integer also it can be noted $k$ and $k+1$ are consecutive integers implying that $k(k+1)=2N$ for some integer $N$ . Putting back we have then $(2k+1)^2\mod(8) \implies (8N +1) \mod(8)\implies (8N+1)^4 \mod(8)=1$ Hence we have then $a^8+b^8+c^8 \equiv 1+1+1\mod (8)\equiv 3\mod(8)$

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Using binomial expansion,

$(2n+1)^8=a+b(2n)+c(2n)^2 + d(2n)^3.....$

Where a, b, c,... are binomial coefficients.

All the terms after the third will be divisible by 8, as all will have 2 to the power $>or=3$ .

b=8, c will be even, which tells us that even the first 2 terms will be divisible by 8.

Which only leaves 1, thus any odd number to the eighth power when divided by 8 leaves the remainder = 1. Thus, three of these numbers will give the remainder as 3 .