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show that $3^{1974} + 5^{1974} \equiv 0 \bmod 13$

My attempt with this question was to use Fermate Little's THM. But I do not understand how to properly use it for this question. Can some one show me a proof.

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    $\begingroup$ What is $1974\pmod {12}$? $\endgroup$ – abiessu Apr 13 '16 at 18:49
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You don't need to use FLT.

\begin{align} 3^{1974}+5^{1974}&\equiv (3^3)^{658} + (5^2)^{987} \pmod{13}\\ &\equiv (27)^{658} + (25)^{987} \pmod{13}\\ &\equiv 1^{658} + (-1)^{987} \pmod{13}\\ &\equiv 1 -1 \pmod{13}\\ &\equiv 0 \pmod{13}\\ \end{align}

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  • $\begingroup$ I prefer the orthography flt for the little one. :-) $\endgroup$ – Brian Tung Apr 13 '16 at 18:52

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