MathLabs

Problem 4

Determine all positive integers relatively prime to all the terms of the infinite sequence an=2n+3n+6n−1, n≥1.a_n=2^n+3^n+6^n -1,\ n\geq 1.
Step 4 of 5: The reciprocals of 2,3,62,3,6 add up to exactly 11
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In plain words

A small numeric coincidence — 12,13,16\frac12,\frac13,\frac16 sum to 11 — is exactly what makes ap−2a_{p-2} vanish modulo every large prime.

12+13+16=1 ⟹ ap−2≡0(modp)\frac12+\frac13+\frac16=1\ \Longrightarrow\ a_{p-2}\equiv0\pmod p
Detailed analysis

Working over the rationals, 12+13+16=3+2+16=1\frac12+\frac13+\frac16=\frac{3+2+1}{6}=1, and the same relation holds among the modular inverses since 2⋅3⋅6=362\cdot3\cdot6=36 is coprime to p>3p>3. So by the previous step ap−2≡(2−1+3−1+6−1)−1≡1−1≡0(modp)a_{p-2}\equiv\left(2^{-1}+3^{-1}+6^{-1}\right)-1\equiv1-1\equiv0\pmod p, i.e. p∣ap−2p\mid a_{p-2}.