When considering $\bmod {36}$, every positive integer $n$ can be written in one of the forms: :$2 k, 4 k + 1, 3 k + 1, 12 k + 11, 18 k + 15, 36 k + 27, 9 k + 3$ $\begin{array}{|c|c|} \hline n \bmod {36} & \text { Form } \\ \hline  0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34 & 2 k \\ \hline  1, 5, 9, 13, 17, 21, 25, 29, 33 & 4 k + 1 \\ \hline  1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34 & 3 k + 1 \\ \hline  11, 23, 35 & 12 k + 11 \\ \hline  15, 33 & 18 k + 15 \\ \hline  27 & 36 k + 27 \\ \hline  3, 12, 21, 30 & 9 k + 3\\ \hline \end{array}$ As seen by inspection, every number less than $36$ is included in the table. And we have: {{begin-eqn}} {{eqn | l = 78557 \times 2^{2 k} + 1       | o = \equiv        | r = 78557 \times 1^k + 1       | rr = \pmod 3       | c = Fermat's Little Theorem and Congruence of Powers }} {{eqn | o = \equiv        | r = 2 + 1       | rr = \pmod 3       | c = Congruence of Product }} {{eqn | o = \equiv        | r = 0       | rr = \pmod 3 }} {{eqn | l = 78557 \times 2^{4 k + 1} + 1       | o = \equiv        | r = 78557 \times 1^k \times 2 + 1       | rr = \pmod 5       | c = Fermat's Little Theorem and Congruence of Powers }} {{eqn | o = \equiv        | r = 2 \times 2 + 1       | rr = \pmod 5       | c = Congruence of Product }} {{eqn | o = \equiv        | r = 0       | rr = \pmod 5 }} {{eqn | l = 78557 \times 2^{3 k + 1} + 1       | r = 78557 \times 8^k \times 2 + 1 }} {{eqn | o = \equiv        | r = 78557 \times 1^k \times 2 + 1       | rr = \pmod 7       | c = Congruence of Powers }} {{eqn | o = \equiv        | r = 3 \times 2 + 1       | rr = \pmod 7       | c = Congruence of Product }} {{eqn | o = \equiv        | r = 0       | rr = \pmod 7 }} {{eqn | l = 78557 \times 2^{12 k + 11} + 1       | o = \equiv        | r = 78557 \times 1^k \times 2^{11} + 1       | rr = \pmod {13}       | c = Fermat's Little Theorem and Congruence of Powers }} {{eqn | o = \equiv        | r = 11 \times 2048 + 1       | rr = \pmod {13}       | c = Congruence of Product }} {{eqn | o = \equiv        | r = 0       | rr = \pmod {13} }} {{eqn | l = 78557 \times 2^{18 k + 15} + 1       | o = \equiv        | r = 78557 \times 1 \times 2^{15} + 1       | rr = \pmod {19}       | c = Fermat's Little Theorem and Congruence of Powers }} {{eqn | o = \equiv        | r = 11 \times 32768 + 1       | rr = \pmod {19}       | c = Congruence of Product }} {{eqn | o = \equiv        | r = 0       | rr = \pmod {19} }} {{eqn | l = 78557 \times 2^{36 k + 27} + 1       | o = \equiv        | r = 78557 \times 1^k \times 2^{27} + 1       | rr = \pmod {37}       | c = Fermat's Little Theorem and Congruence of Powers }} {{eqn | o = \equiv        | r = 6 \times 512^3 + 1       | rr = \pmod {37}       | c = Congruence of Product }} {{eqn | o = \equiv        | r = 6 \times \paren {-6}^3 + 1       | rr = \pmod {37}       | c = Congruence of Powers }} {{eqn | o = \equiv        | r = 0       | rr = \pmod {37} }} {{eqn | l = 78557 \times 2^{9 k + 3} + 1       | o = \equiv        | r = 78557 \times 512^k \times 2^3 + 1 }} {{eqn | o = \equiv        | r = 78557 \times 1^k \times 8 + 1       | rr = \pmod {73}       | c = Congruence of Powers }} {{eqn | o = \equiv        | r = 9 \times 8 + 1       | rr = \pmod {73}       | c = Congruence of Product }} {{eqn | o = \equiv        | r = 0       | rr = \pmod {73} }} {{end-eqn}} Therefore $78557 \times 2^n + 1$ is always divisible by one of $\set {3, 5, 7, 13, 19, 37, 73}$. Hence they are composite for every $n$. {{qed}}