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$F_n$ is the $n$-th Fermat Number. $F_n = 2^{2^n} + 1$. Prove there are infinitely many values of n for which $F_n + 2$ is composite. I tried using reduce modulo 7 but got stuck. Any help is appreciated!

Bill Dubuque
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Emma Johnson
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    I think your idea is solid. Can you write out the first few values for $2^{(2^n)}+3\pmod 7$? – lulu May 17 '20 at 01:29
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    Show $2^n\equiv2\bmod6$ for all odd $n$, and also $2^{6k+2}+3\equiv0\bmod7$ for all $k$. – Gerry Myerson May 17 '20 at 03:43
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    The usual notation for Fermat numbers has them start with $F_0=3$, so $F_n$ is the $(n+1)^\text{st}$ Fermat number, not the $n^\text{th}$. Anyway, to show that the Fermat numbers are alternately $3$ and $5$ modulo $7$, just use the defining recurrence $$F_n=F_0F_1\cdots F_{n-1}+2$$ and the fact that $3\cdot5\equiv1\pmod7$. – bof May 17 '20 at 06:23
  • In case you haven't seen that recursive definition of $F_n$, see this question: https://math.stackexchange.com/questions/2387336/induction-on-fermat-numbers-f-n-prod-j-0n-1f-j2 – bof May 17 '20 at 06:33

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Note that $$F_{n+1}-1=(F_n-1)^2,$$ that is, $$F_{n+1}=(F_n-1)^2+1.$$ Now $$F_0=3,$$ and $$F_n\equiv3\pmod7\implies F_{n+1}\equiv5\pmod7,$$ and $$F_n\equiv5\pmod7\implies F_{n+1}\equiv17\equiv3\pmod7.$$ It follows that $F_n\equiv5\pmod7$ for $n=1,3,5,7,\dots$, and $F_n+2$ is composite for $n=3,5,7,\dots$.

Alternatively, you could use the recurrence $$F_{n+2}=(F_n-1)^4+1.$$

bof
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