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# Divisibility by 16 (Tomato Subjective 35)

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

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

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

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August 28, 2013

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