Problem 49 **

Show that 7 \mid \left(5555^{2222} + 2222^{5555}\right).

[originally posted by Llwellyn on TSR]

5555^{2222}+2222^{5555}=5^{2222}(11\times 101)^{2222}+2^{5555} (11\times 101)^{5555}
\hspace{3.9cm}=(11\times 101)^{2222}\left [ 5^{2222}+2^{5555} (11 \times 101)^{3333} \right ]

Clearly 7\nmid(11\times 101)^{2222} \therefore it suffices to show that 7\mid5^{2222}+2^{5555} (11 \times 101)^{3333}

5^{2222}=(7-2)^{2222}\equiv 2^{2222}(\mod7)

\therefore \left [5^{2222}+2^{5555} (11 \times 101)^{3333} \right ] (\mod7)\equiv \left[ 2^{2222} +2^{5555} (11 \times 101)^{3333}\right] (\mod7)
\hspace{8.12cm}=\left[2^{2222}(1+2222^{3333})\right](\mod7)

If we can show that (1+2222^{3333})\equiv0(\mod7) we are done.

2222^{3333}=(7\times 317+3)^{3333}=7k+3^{3333}\qquad k\in\mathbb{Z}
\hspace{5.65cm}=7k+27^{1111}
\hspace{5.65cm}=7k+(4\times7 -1)^{1111}
\hspace{5.65cm}=7k+7m+(-1)^{1111}\qquad m\in\mathbb{Z}
\hspace{5.65cm}=7k+7m-1
\therefore 1+2222^{3333}=7(m+k)\equiv0(\mod7)

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