We check the result and show that it cannot be improved further by showing: :$3906$ is the smallest $x$ such that $x^6 + 1091$ is prime. Suppose $x^6 + 1091$ is prime. Then: :$x$ is a multiple of $42$ :$x$ ends in $0$, $4$ or $6$ in decimal notation :$x \not \equiv \pm 1, \pm 3, \pm 4 \pmod {13}$ :$x \not \equiv \pm 4, \pm 6, \pm 9 \pmod {19}$ The proof is split into $6$ parts: === $x$ is a multiple of $2$ === Suppose not. Then $x$ is odd, and so is $x^6$. Hence $x^6 + 1091$ is even, and thus composite. Thus we must require $x$ to be even. {{qed|lemma}} === $x$ is a multiple of $3$ === Suppose not. Write $x = 3 k \pm 1$. Hence: {{begin-eqn}} {{eqn | l = x^6 + 1091       | r = \paren {3 k \pm 1}^6 + 1091 }} {{eqn | o = \equiv       | r = \paren {\pm 1}^6 - 1       | rr = \pmod 3       | c = Congruence of Powers }} {{eqn | o = \equiv       | r = 0       | rr = \pmod 3 }} {{end-eqn}} showing that $x^6 + 1091$ is composite (divisible by $3$). Thus we must require $x$ to be a multiple of $3$. {{qed|lemma}} === $x$ is a multiple of $7$ === Suppose not. Then $x \perp 7$. Then: {{begin-eqn}} {{eqn | l = n^6 + 1091       | o = \equiv       | r = 1 + 1091       | rr = \pmod 7       | c = Fermat's Little Theorem }} {{eqn | o = \equiv       | r = 0       | rr = \pmod 7 }} {{end-eqn}} showing that $x^6 + 1091$ is composite (divisible by $7$). Thus we must require $x$ to be a multiple of $7$. {{qed|lemma}} === $x$ ends in $0$, $4$ or $6$ === From the above we require $x$ to be even. Suppose $x$ ends in $2$ or $8$. Then $x^6$ ends in $4$. Thus $x^6 + 1091$ ends in $5$, which by Divisibility by 5 is divisible by $5$. Hence we must require $x$ to end in $0$, $4$ or $6$. {{qed|lemma}} === $x \not \equiv \pm 1, \pm 3, \pm 4 \pmod {13}$ === Here is a table of $x^6 \pmod {13}$: :$\begin{array}{|c|c|c|c|c|c|c|c|} \hline  x \bmod {13} & 0 & \pm 1 & \pm 2 & \pm 3 & \pm 4 & \pm 5 & \pm 6 \\ \hline x^6 \bmod {13} & 0 & 1 & -1 & 1 & 1 & -1 & -1 \\ \hline \end{array}$ For $x \equiv \pm 1, \pm 3, \pm 4 \pmod {13}$: {{begin-eqn}} {{eqn | l = x^6 + 1091       | o = \equiv       | r = 1 - 1       | rr = \pmod {13} }} {{eqn | o = \equiv       | r = 0       | rr = \pmod {13} }} {{end-eqn}} showing that $x^6 + 1091$ is composite (divisible by $13$). Thus we must require $x \not \equiv \pm 1, \pm 3, \pm 4 \pmod {13}$. {{qed|lemma}} === $x \not \equiv \pm 4, \pm 6, \pm 9 \pmod {19}$ === Here is a table of $x^6 \pmod {19}$: :$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline  x \bmod {19} & 0 & \pm 1 & \pm 2 & \pm 3 & \pm 4 & \pm 5 & \pm 6 & \pm 7 & \pm 8 & \pm 9 \\ \hline x^6 \bmod {19} & 0 & 1 & 7 & 7 & 11 & 6 & 11 & 1 & 1 & 11\\ \hline \end{array}$ For $x \equiv \pm 4, \pm 6, \pm 9 \pmod {19}$: {{begin-eqn}} {{eqn | l = x^6 + 1091       | o = \equiv       | r = 11 + 8       | rr = \pmod {19} }} {{eqn | o = \equiv       | r = 0       | rr = \pmod {19} }} {{end-eqn}} showing that $x^6 + 1091$ is composite (divisible by $19$). Thus we must require $x \not \equiv \pm 4, \pm 6, \pm 9 \pmod {19}$. {{qed|lemma}} Here is a list of $x^6 + 1091$, where $42 \divides x$ and does not end in $2$ or $8$: {{begin-eqn}} {{eqn | l = 84^6 + 1091       | r = 23 \times 15273827509 }} {{eqn | l = 126       | o = \equiv       | r = -4 \pmod {13} }} {{eqn | l = 210^6 + 1091       | r = 109 \times 786845146799 }} {{eqn | l = 294       | o = \equiv       | r = +9 \pmod {19} }} {{eqn | l = 336       | o = \equiv       | r = -6 \pmod {19} }} {{eqn | l = 420       | o = \equiv       | r = +4 \pmod {13} }} {{eqn | l = 504       | o = \equiv       | r = -3 \pmod {13} }} {{eqn | l = 546^6 + 1091       | r = 2129 \times 12444578592403 }} {{eqn | l = 630^6 + 1091       | r = 347 \times 41957 \times 4294468229 }} {{eqn | o = etc.}} {{end-eqn}} {{finish|1/8 of the way there. I'll fill it in later}} The above results are obtained using an integer factorization calculator. {{qed}}