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  "body": "\n## Proof\n\nLet $n \\ge 1$ and $F = F_n = 2^{2^n} + 1$, a [[Fermat number|0R03B54V4Z]]. Put $e = (F - 1)/2 = 2^{2^n - 1}$. Congruences are [[modulo|0TPW25CBN6]] $F$ unless stated.\n\n**If $F$ is prime, then $3^e \\equiv -1$.**\n\n1. Since $n \\ge 1$, we have $2^n \\ge 2$, so $4$ divides $2^{2^n}$. Hence $F \\equiv 1 \\pmod 4$, and $F \\ge 5$.\n2. Also $2^{2^n} = 4^{2^{n-1}} \\equiv 1 \\pmod 3$. Hence $F \\equiv 2 \\pmod 3$.\n3. Assume $F$ is [[prime|0SN0MAEKYX]]. Then $F$ and $3$ are distinct odd primes.\n4. By the [[law of quadratic reciprocity|085MJS6907]], with [[Legendre symbols|01J8AMV9RS]], $\\left(\\frac{3}{F}\\right)\\left(\\frac{F}{3}\\right) = (-1)^{1 \\cdot e} = 1$, because $e$ is even.\n5. The squares modulo $3$ are $0^2 \\equiv 0$, $1^2 \\equiv 1$ and $2^2 \\equiv 1$. So $F \\equiv 2$ is a [[quadratic nonresidue|0SQDB87Z3T]] modulo $3$, and $\\left(\\frac{F}{3}\\right) = -1$.\n6. By step 4, $\\left(\\frac{3}{F}\\right) = -1$.\n7. By [[Euler's criterion|0M7DEDKSYB]], $3^e \\equiv \\left(\\frac{3}{F}\\right) = -1 \\pmod F$.\n\n**If $3^e \\equiv -1$, then $F$ is prime.**\n\n8. Assume $3^e \\equiv -1 \\pmod F$. A common divisor of $3$ and $F$ divides $3^e$ and $3^e + 1$, hence divides $1$. So $\\gcd(3, F) = 1$.\n9. So the class $[3]$ lies in the group $G = (\\Z/F\\Z)^\\times$ of [[units|05THZKR9AN]], by [[the units of Z/nZ|0GXNHEHQNA]]. Let $r$ be its order in $G$.\n10. Squaring the assumption gives $3^{2e} = 3^{F-1} \\equiv 1$. By [[the order criterion|0F5QC101Q7]], $r$ divides $F - 1 = 2^{2^n}$. So $r = 2^j$ with $0 \\le j \\le 2^n$.\n11. If $j \u003c 2^n$, then $r$ divides $e = 2^{2^n - 1}$, so $3^e \\equiv 1$. With $3^e \\equiv -1$ this gives $F \\mid 2$, which is false since $F \\ge 5$.\n12. So $r = 2^{2^n} = F - 1$.\n13. The [[order of an element divides the order of the group|08V7JHS3QZ]]. The group $G$ has $\\varphi(F)$ elements, by [[the order of the units group|0S7495H4HM]]. So $F - 1 \\le \\varphi(F)$.\n14. By the definition of [[Euler's totient function|0VV04M41AN]], $\\varphi(F)$ counts integers $k$ with $1 \\le k \\le F$ coprime to $F$. The integer $k = F$ is not coprime to $F \u003e 1$, so $\\varphi(F) \\le F - 1$.\n15. Hence $\\varphi(F) = F - 1$, and each $k$ with $1 \\le k \\le F - 1$ is coprime to $F$.\n16. If $d$ were a divisor of $F$ with $1 \u003c d \u003c F$, then $\\gcd(d, F) = d \u003e 1$, contrary to step 15. So $F$ is prime.\n\n## Idea\n\nIf $F_n$ is prime, quadratic reciprocity shows that $3$ is not a square modulo $F_n$, and Euler's criterion turns this into the congruence. Conversely, the congruence makes the order of $3$ equal to $F_n - 1$. Only a prime modulus has that many units.\n",
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