We have that: {{begin-eqn}} {{eqn | l = 3^{91}       | r = 26 \, 183 \, 890 \, 704 \, 263 \, 137 \, 277 \, 674 \, 192 \, 438 \, 430 \, 182 \, 020 \, 124 \, 347       | c =  }} {{eqn | r = 26 \, 183 \, 890 \, 704 \, 263 \, 137 \, 277 \, 674 \, 192 \, 438 \, 430 \, 182 \, 020 \, 124 \, 344 + 3       | c =  }} {{eqn | r = 91 \times 287 \, 735 \, 062 \, 684 \, 210 \, 299 \, 754 \, 661 \, 455 \, 367 \, 364 \, 637 \, 583 \, 784 + 3       | c =  }} {{eqn | ll= \leadsto       | l = 3^{91}       | o = \equiv       | r = 3 \pmod {91}       | c =  }} {{end-eqn}} Here we define Fermat pseudoprime to base $3$ to be a composite number $n$ such that (a slightly stronger result): :$3^{n - 1} \equiv 1 \pmod n$ Otherwise we have: :$3^6 = 729 \equiv 3 \pmod 6$ Following our definition, we see that a Fermat pseudoprime to base $3$ cannot be divisible by $3$: {{begin-eqn}} {{eqn | l = 3^{3 k - 1}       | o = \equiv       | r = 0       | rr = \pmod 3 }} {{eqn | o = \not \equiv       | r = 1       | rr = \pmod {3 k} }} {{end-eqn}} neither can it be divisible by $4$: {{begin-eqn}} {{eqn | l = 3^{4 k}       | r = \paren {3^{2 k} }^2 }} {{eqn | o = \equiv       | r = 1       | rr = \pmod 4       | c = Square Modulo 4 }} {{eqn | o = \not \equiv       | r = 3       | rr = \pmod {4 k} }} {{end-eqn}} nor $10$: {{begin-eqn}} {{eqn | l = 3^{10 k}       | r = \paren {3^{5 k} }^2 }} {{eqn | o = \equiv       | r = \pm 1       | rr = \pmod 5       | c = Square Modulo 5 }} {{eqn | o = \not \equiv       | r = 3       | rr = \pmod {10 k} }} {{end-eqn}} Also such a number cannot be equal to twice a prime greater than $3$: {{begin-eqn}} {{eqn | l = 3^{2 p}       | o = \equiv       | r = 3^2       | rr = \pmod p       | c = Fermat's Little Theorem }} {{eqn | o = \not \equiv       | r = 3       | rr = \pmod p       | c = Since $p \nmid 6 = 2 \times 3$ }} {{eqn | ll = \leadsto       | l = 3^{2 p}       | o = \not \equiv       | r = 3       | rr = \pmod {2 p} }} {{end-eqn}} nor five times a prime greater than $5$: {{begin-eqn}} {{eqn | l = 3^{5 p}       | o = \equiv       | r = 3^5       | rr = \pmod p       | c = Fermat's Little Theorem }} {{eqn | o = \not \equiv       | r = 3       | rr = \pmod p       | c = Since $p \nmid 240 = 2^4 \times 3 \times 5$ }} {{eqn | ll = \leadsto       | l = 3^{5 p}       | o = \not \equiv       | r = 3       | rr = \pmod {5 p} }} {{end-eqn}} Therefore only the following numbers less than $91$ remain: :$25, 49, 77$ and we check: {{begin-eqn}} {{eqn | l = 3^{25}       | r = 847 \, 288 \, 609 \, 443       | c =  }} {{eqn | r = 33 \, 891 \, 544 \, 377 \times 25 + 18       | c =  }} {{eqn | ll= \leadsto       | l = 3^{25}       | o = \equiv       | r = 18 \pmod {25}       | c =  }} {{eqn | l = 3^{49}       | r = 239 \, 299 \, 329 \, 230 \, 617 \, 529 \, 590 \, 083       | c =  }} {{eqn | r = 4 \, 883 \, 659 \, 780 \, 216 \, 684 \, 277 \, 348 \times 49 + 31       | c =  }} {{eqn | ll= \leadsto       | l = 3^{49}       | o = \equiv       | r = 31 \pmod {49}       | c =  }} {{eqn | l = 3^{77}       | r = 5 \, 474 \, 401 \, 089 \, 420 \, 219 \, 382 \, 077 \, 155 \, 933 \, 569 \, 751 \, 763       | c =  }} {{eqn | r = 71 \, 096 \, 118 \, 044 \, 418 \, 433 \, 533 \, 469 \, 557 \, 578 \, 827 \, 944 \times 77 + 75       | c =  }} {{eqn | ll= \leadsto       | l = 3^{77}       | o = \equiv       | r = 75 \pmod {77}       | c =  }} {{end-eqn}} hence the result. {{qed}}