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<strong>(a) Prove that, for any odd integer n, $n^4$ when divided by 16 always leaves remainder 1.</strong>

<strong>(b) Hence or otherwise show that we cannot find integers $n_1 , n_2 , ... , n_8$ such that $n_1^4 + n_2^4 + ... + n_8^4 = 1993$.</strong>

Solution: For part (a) we consider n = 2k +1 and expand it’s fourth power binomially to get $(2k+1)^4 = \binom {4}{0} (2k)^4 +\binom {4}{1} (2k)^3 + \binom {4}{2} (2k)^2 + \binom {4}{3} (2k)^1 + \binom {4}{4} (2k)^0 = 16k^4 + 32k^3 + 24k^2 + 8k + 1$

Now $24k^2 + 8k = 8k(3k+1)$ ; if k is even then 8k is divisible by 16 and if k is odd 3k+1 is even and product of 8k and 3k+1 is divisible by 16. Since $16k^4 + 32 k^3$ is already divisible by 16 we conclude $(2k+1)^4$ when divided by 16 gives 1 as remainder.

For part (b) we note that 1993 when divided by 16 produces 9 as the remainder. Each of the eight of fourth powers when divided by 16 produces either 0 (when $n_i$ is even) or 1 (when $n_i$ is odd using part (a)) as remainder. Thus they can add up to at most 8 (modulo 16) hence can never be equal to 9 (which 1993 is modulo 16).
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