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I am getting into a bit of number theory and came up with the following exercise:

Exercise. Show that each Fermat number is prime iff $F_n | 3^{\frac{F_n-1}{2}}+1$.

My only idea follows (it is not a complete resolution).

Idea. If $F_n | 3^{\frac{F_n-1}{2}}+1$ then $$3^{\frac{F_n-1}{2}}+1 \equiv 0 \hspace{.3cm}(\text{mod} F_n) \Leftrightarrow 3^{\frac{F_n-1}{2}} \equiv -1 \hspace{.3cm}(\text{mod} F_n)$$ My next step was to consider $F_n = 2^{2^n}+1$. This means that $\frac{F_n-1}{2} = 2^{2^n-1}$ and thus the above is equivalent to: $$ 3^{2^{2^n-1}} \equiv -1 \hspace{.3cm} (\text{mod} F_n)$$ But now I don't know how to proceed further on. I have thought about raising the congruence to some exponent (this technique can be seen here: Show that every composite Fermat number is a pseudoprime base 2.) but haven't been able to came up with a solution.

Thanks for any help in advance.

xyz
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    This is called Pepin's test. Searching on that name will easily locate proofs - as will searching on "fermat number" "primality test". One should always search before posting questions. – Bill Dubuque Jul 15 '22 at 19:03
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    @BillDubuque you'd be out of a job if we did. – Roddy MacPhee Jul 15 '22 at 19:58
  • @BillDubuque Thanks for the help. English is not my native language so sometimes it's hard to find the correct words to search for. Roddy got some jokes ;) ! – xyz Jul 15 '22 at 21:02
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    @roro This "joke" is similar to the joke that doctors would have no work if noone would get ill. In this case, I clearly agree Bill since this is a very famous criterion for Fermat numbers that can easily be googled. – Peter Jul 16 '22 at 09:31
  • I agree with Bill aswell! He is absolutely right, I just found some fun about Roddy's comment :D – xyz Jul 16 '22 at 09:34

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