Numbers for which Sixth Power plus 1091 is Composite

Theorem
The number $1091$ has the property that:
 * $x^6 + 1091$

is composite for all integer values of $x$ from $1$ to $3905$.

Proof
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.

$x$ is a multiple of $3$
Suppose not. Write $x = 3 k \pm 1$.

Hence:

showing that $x^6 + 1091$ is composite (divisible by $3$).

Thus we must require $x$ to be a multiple of $3$.

$x$ is a multiple of $7$
Suppose not. Then $x \perp 7$.

Then:

showing that $x^6 + 1091$ is composite (divisible by $7$).

Thus we must require $x$ to be a multiple of $7$.

$x$ 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 $6$.

$x \not \equiv \pm 1, \pm 3, \pm 4 \pmod {13}$
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 $x \equiv \pm 1, \pm 3, \pm 4 \pmod {13}$:

showing that $x^6 + 1091$ is composite (divisible by $13$).

Thus we must require $x \not \equiv \pm 1, \pm 3, \pm 4 \pmod {13}$.

$x \not \equiv \pm 4, \pm 6, \pm 9 \pmod {19}$
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 $x \equiv \pm 4, \pm 6, \pm 9 \pmod {19}$:

showing that $x^6 + 1091$ is composite (divisible by $19$).

Thus we must require $x \not \equiv \pm 4, \pm 6, \pm 9 \pmod {19}$.

Here is a list of $x^6 + 1091$, where $42 \divides x$ and does not end in $2$ or $8$:

The above results are obtained using an integer factorization calculator.