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  "body": "\n## Proof\n\nLet $p$ be an odd prime and $M = M_p = 2^p - 1$, the [[Mersenne number|0GJBJ19NXR]]. Then $p \\ge 3$, so $M \\ge 7$ and $M$ is odd.\n\n**Setup.**\n\n1. Let $A = \\set{a + b\\sqrt{3} : a, b \\in \\Z}$. It contains $1$ and is closed under differences and under products, since $(a + b\\sqrt3)(c + d\\sqrt3) = (ac + 3bd) + (ad + bc)\\sqrt3$.\n2. So $A$ is a [[subring|0859D58WPD]] of $\\R$, and it is a [[commutative ring|0GYGHKX834]].\n3. If $a + b\\sqrt3 = c + d\\sqrt3$ with integers $a, b, c, d$ and $b \\ne d$, then $\\sqrt3 = (a - c)/(d - b)$ is rational. This contradicts [[the irrationality of square roots of non-squares|0N10YRNNWV]]. So $a = c$ and $b = d$.\n4. For an integer $N \\ge 2$, the set $NA = \\set{N\\alpha : \\alpha \\in A}$ is a [[two-sided ideal|0DDA6CY7GJ]] of $A$. Let $X_N = A/NA$ be the [[quotient ring|0VFW4Y69BT]]. We write $\\bar\\alpha$ for the coset of $\\alpha$.\n5. Let $s$ be an integer. If $s \\in NA$, then $s = Nc + Nd\\sqrt3$ for integers $c, d$. By step 3, $Nd = 0$ and $s = Nc$. So $\\bar s = 0$ in $X_N$ exactly when $N \\mid s$.\n6. Put $\\omega = 2 + \\sqrt3$ and $\\tau = 2 - \\sqrt3$ in $A$. Then $\\omega\\tau = 4 - 3 = 1$.\n7. We show $s_i = \\omega^{2^i} + \\tau^{2^i}$ by [[induction|0ASJVEFA9J]] on $i$. For $i = 0$, $\\omega + \\tau = 4 = s_0$.\n8. If $s_{i-1} = \\omega^{2^{i-1}} + \\tau^{2^{i-1}}$, then $s_{i-1}^2 - 2 = \\omega^{2^i} + 2(\\omega\\tau)^{2^{i-1}} + \\tau^{2^i} - 2 = \\omega^{2^i} + \\tau^{2^i}$, by step 6.\n9. Put $t = 2^{p-2}$. So $s_{p-2} = \\omega^t + \\tau^t$, and $2t = 2^{p-1} = (M + 1)/2$.\n\n**If $s_{p-2} \\equiv 0 \\pmod M$, then $M$ is prime.**\n\n10. Assume $M \\mid s_{p-2}$ and, for a contradiction, that $M$ is [[composite|0N3ADXHNTQ]]. By [[the square root bound|0EB6BBMQMC]], $M$ has a prime factor $q$ with $q^2 \\le M$. Since $M$ is odd, $q$ is odd.\n11. Work in $X = X_q$. Since $q \\mid s_{p-2}$, step 5 gives $\\bar\\omega^t + \\bar\\tau^t = 0$.\n12. Multiply by $\\bar\\omega^t$ and use $\\omega\\tau = 1$. This gives $\\bar\\omega^{2t} + 1 = 0$, so $\\bar\\omega^{2^{p-1}} = -1$ and $\\bar\\omega^{2^p} = 1$.\n13. In $X$ we have $-1 \\ne 1$: otherwise $\\bar 2 = 0$, so $q \\mid 2$ by step 5, but $q$ is odd.\n14. By step 6, $\\bar\\omega$ is a [[unit|019NFGFPR5]] of $X$, so it lies in the [[group of units|05THZKR9AN]] $X^\\times$. Let $r$ be its [[order|0WE2NHZJ7B]].\n15. By step 12 and [[the order criterion|0F5QC101Q7]], $r$ divides $2^p$, so $r = 2^j$ with $j \\le p$. If $j \\le p - 1$, then $\\bar\\omega^{2^{p-1}} = 1$, against steps 12 and 13. So $r = 2^p$.\n16. Each element of $A$ is congruent modulo $qA$ to some $a + b\\sqrt3$ with $0 \\le a, b \u003c q$. So $X$ has at most $q^2$ elements.\n17. The element $0$ of $X$ is not a unit, since $0 \\cdot \\beta = 0 \\ne 1$ by step 13. So $X^\\times$ has at most $q^2 - 1$ elements.\n18. Since the [[order of an element divides the order of the group|08V7JHS3QZ]], $2^p \\le q^2 - 1 \u003c q^2 \\le M = 2^p - 1$. This is false. So $M$ is prime.\n\n**If $M$ is prime, then $s_{p-2} \\equiv 0 \\pmod M$.**\n\n19. Assume $M$ is [[prime|0SN0MAEKYX]]. Congruences below are [[modulo|0TPW25CBN6]] $M$. Since $2^p \\equiv 1$ and $p$ is odd, $2 \\equiv 2^{p+1} = (2^{(p+1)/2})^2$.\n20. So $2$ is a [[quadratic residue|0SQDB87Z3T]] modulo $M$, and $M \\nmid 2$. By [[Euler's criterion|0M7DEDKSYB]], $2^{(M-1)/2} \\equiv 1$.\n21. Since $p$ is odd, $2^p = 2 \\cdot 4^{(p-1)/2} \\equiv 2 \\pmod 3$. So $M \\equiv 1 \\pmod 3$, and $M \\ne 3$.\n22. The number $1 = 1^2$ is a quadratic residue modulo $3$, so the [[Legendre symbol|01J8AMV9RS]] $\\left(\\frac{M}{3}\\right)$ equals $1$.\n23. Also $(M - 1)/2 = 2^{p-1} - 1$ is odd. By [[quadratic reciprocity|085MJS6907]], $\\left(\\frac{3}{M}\\right)\\left(\\frac{M}{3}\\right) = (-1)^{(M-1)/2} = -1$. So $\\left(\\frac{3}{M}\\right) = -1$.\n24. By Euler's criterion, $3^{(M-1)/2} \\equiv -1$.\n25. For $0 \u003c k \u003c M$, the [[factorial formula|033KTKKRV6]] gives $k!\\,(M - k)! \\binom{M}{k} = M!$. The prime $M$ divides $M!$ but no factor of $k!\\,(M-k)!$. By [[Euclid's lemma|051VYKA1EX]], $M \\mid \\binom{M}{k}$.\n26. Work in $X = X_M$. For integers $a, b$, the [[binomial theorem|0EHWPWGHB2]] and step 25 give $\\overline{(a + b\\sqrt3)^M} = \\overline{a^M + b^M (\\sqrt3)^M}$. Here $(\\sqrt3)^M = 3^{(M-1)/2}\\sqrt3$, since $M$ is odd.\n27. By [[Fermat's little theorem|0BMXWXDVJZ]] and step 24, $a^M \\equiv a$ and $b^M 3^{(M-1)/2} \\equiv -b$. So $\\overline{(a + b\\sqrt3)^M} = \\overline{a - b\\sqrt3}$.\n28. Note $(1 + \\sqrt3)^2 = 4 + 2\\sqrt3 = 2\\omega$. With step 27 for $a = b = 1$,\n$$\\overline{(2\\omega)^{(M+1)/2}} = \\overline{(1 + \\sqrt3)^{M+1}} = \\overline{(1 + \\sqrt3)(1 - \\sqrt3)} = \\overline{-2}.$$\n29. By step 20, $\\overline{2^{(M+1)/2}} = \\overline{2 \\cdot 2^{(M-1)/2}} = \\bar 2$. So $\\bar 2 \\, \\bar\\omega^{(M+1)/2} = -\\bar 2$.\n30. The element $\\bar 2$ is a unit of $X$, with inverse $\\overline{(M+1)/2}$. Hence $\\bar\\omega^{(M+1)/2} = -1$, that is, $\\bar\\omega^{2t} = -1$ by step 9.\n31. Multiply by $\\bar\\tau^t$ and use $\\omega\\tau = 1$. This gives $\\bar\\omega^t = -\\bar\\tau^t$, so $\\overline{s_{p-2}} = \\bar\\omega^t + \\bar\\tau^t = 0$.\n32. By step 5, $M$ divides $s_{p-2}$. So $s_{p-2} \\equiv 0 \\pmod{M_p}$.\n\n## Idea\n\nThe sequence is $s_i = \\omega^{2^i} + \\omega^{-2^i}$ with $\\omega = 2 + \\sqrt3$. If $M_p$ divides $s_{p-2}$, then $\\omega$ has order $2^p$ modulo any prime factor $q$, which is too large when $q^2 \\le M_p$. If $M_p$ is prime, $3$ is a nonresidue, so the $M_p$-th power map acts like conjugation on $a + b\\sqrt3$. This forces $\\omega^{(M_p+1)/2} = -1$.\n",
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