I am asked to prove that $3^{2n}+7$ is divisible by 8, for all positive integers, I have prove that: For $n=1$, $3^{2(1)} +7=16$, then for $n=k$, $3^{2k}+7=8P$, where P is a positive integer and for $n=k+1$, $3^{2k+3}+7=8P$. I have no further development and dont know how to continue...¿What should i do from here to prove?
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Is it $3^{2n}$ or $3^2n$? – ajotatxe Oct 07 '22 at 01:05
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Sorry, i just edited the question – Yocheved_Vered Oct 07 '22 at 01:11
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1Does this answer your question? Prove that $3^{2n} +7$ is divisible by 8 - found using an Approach0 search. – John Omielan Oct 07 '22 at 01:22
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@Yocheved_Vered Welcome to Math SE. Note the search also turned up the similar Use induction to "establish" the divisibility statement $8 | 5^{2n}+7$, which is closed as a duplicate of Mathematical Induction divisibility $8\mid 3^{2n}-1$, with this also basically being a duplicate of your question since $3^{2n}-1=(3^{2n}+7)-8$, so $8\mid 3^{2n}-1 \iff 8\mid 3^{2n}+7$. – John Omielan Oct 07 '22 at 01:26
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You've shown the claim holds for $n = 1$. Now suppose the claim holds for some $n \in \mathbb{N}$ and show that this implies that it also holds for $n + 1$. So, for $n + 1$ we have
$$3^{2(n + 1)} + 7 = 3^{2n}3^2 + 7 = 9 \cdot 3^{2n} + 7.$$
Now find a way to rewrite $\left ( 9 \cdot 3^{2n} \right ) + 7$ that allows you to use the induction hypothesis to show that this number is divisible by 8.
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