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Number theory Difficulty 4.5 AIME Prove it North Macedonia

Let nn be a natural number. Prove that 137n+n513|7^n + n^5 if and only if 13n77n+113|n^7 \cdot 7^n + 1.

Solution

Let 137n+n513|7^n + n^5. Clearly 1313 does not divide nn, so by Little Fermat's theorem we have that n121(mod13)n^{12} \equiv 1 \pmod{13}, i.e. 13n12113|n^{12} - 1.

Now we have 137n+n513|7^n + n^5, i.e. 13n7(7n+n5)13|n^7(7^n + n^5) or 13n77n+1+n12113|n^7 7^n + 1 + n^{12} - 1, and because of 13n12113|n^{12} - 1 we get 13n77n+113|n^7 \cdot 7^n + 1.

Now let 13n77n+113|n^7 \cdot 7^n + 1; clearly 1313 does not divide nn, so as previously we have 13n12113|n^{12} - 1. We get 13n77n+113|n^7 \cdot 7^n + 1, i.e. 13n5(n77n+1)13|n^5(n^7 \cdot 7^n + 1), i.e. 13n127n+n513|n^{12} \cdot 7^n + n^5, or 13(n121)7n+7n+n513|(n^{12} - 1)7^n + 7^n + n^5, from where we have that 137n+n513|7^n + n^5, which concludes the proof.

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