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How can I show this? I was doing some proves, and this question came out of nowhere, but I felt curious about this and so I tried this. The question is that given any $k \in {{\Bbb Z}^ + }$ such that $k \geqslant 5$, show that ${6^k} + 1$ is composite. This question looks and feels extremely difficult to prove for me, and so I want help from this. Luckily, I did some proves such as using By Induction, By Contradiction, and more, and nothing worked. Also, I tried looking this problem through the Internet, and nothing...

What I tried to do was the following, though:

Show that if $$A = \{ {6^k} + 1|{6^k} + 1{\text{ is prime}}\} $$then $A = \{ 7,37,1297\} $. Especially, only when $k = 1$, $k = 2$ or $k = 4$, ${6^k} + 1$ is prime. Or equavilently, ${6^k} + 1$ is prime iff $k = 1$, $k = 2$ or $k = 4$.

Any ideas?

Thanks, Joshua.

Update: This question is already answered. Thanks, everyone for helping.

Update 2: I think that I can prove this (but not completely).

Proof:

Let $k \in {{\Bbb Z}^ + }$ and odd. Then $k = 2m + 1,m \geqslant 2$. $\Rightarrow {6^k} + 1 = {6^{2m + 1}} + 1 = 6 \cdot {36^m} + 1$. Since $36 \equiv 1(\bmod 7) \Rightarrow 6 \cdot {36^m} + 1 \equiv 6 \cdot {1^m} + 1 = 6 \cdot 1 + 1 = 7 \equiv 0(\bmod 7)$. Hence when k is odd, $6^k + 1$ is composite.

If $k = ab$ and $b$ is odd, then $6^k + 1 = 6^{ab} + 1 = 6^{ab} + 1 \Rightarrow 6^a + 1|(6^{ab}+1)$.

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    $\begingroup$ For $k$ odd $6^k+1$ is a multiple of 7. It remains to check what happens if $k$ is even. $\endgroup$
    – cos_dm_math21
    Commented Jun 7, 2021 at 22:06
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    $\begingroup$ Much stronger: If $k$ has an odd divisor $d>1$, say $k = ds$, then $6^k+1$ is a multiple of $6^s+1$. So it remains to check what happens if $k$ is a power of $2$. $\endgroup$
    – cos_dm_math21
    Commented Jun 7, 2021 at 22:10
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    $\begingroup$ Just a remark: It is still an open problem if there exists some $n>4$ such that $F_n = 2^{2^n}+1$ is prime. These numbers are called Fermat's numbers. Here you are dealing with $G_n = 6^{2^n}+1$ (for $k=2^n$) and your claim is that $G_n$ is prime if and only if $n=1, 2, 3$. $\endgroup$
    – cos_dm_math21
    Commented Jun 7, 2021 at 22:22
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    $\begingroup$ @user938668 Welcome to Math SE. FYI, what you're asking about is the Generalized Fermat primes of $F_n(a) = a^{2^n} + 1$ where $a = 6$. The Wikipedia article section says that "The smallest prime number $F_n(a)$ with $n \gt 4$ is $F_5(30)$, or $30^{32} + 1$." Thus, any primes of $6^k + 1$ where $k \gt 4$ would involve relatively large values. $\endgroup$
    – John Omielan
    Commented Jun 7, 2021 at 22:22
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    $\begingroup$ We already have many closely related question on generalized Fermat numbers. If the comments and the answers in the dupe don't answer your question then please elaborate in comments here. $\endgroup$
    – Bill Dubuque
    Commented Jun 7, 2021 at 22:26

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