设 $k$ 是正整数. 证明 $\mathrm{gcd}(k,2^n+3^n+6^n-1)=1$ 对所有正整数 $n$ 成立当且仅当 $k=1$.
设 $k$ 是正整数. 证明 $\mathrm{gcd}(k,2^n+3^n+6^n-1)=1$ 对所有正整数 $n$ 成立当且仅当 $k=1$.
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Let $k$ be a positive integer. Prove that $\mathrm{gcd}(k,2^n+3^n+6^n-1)=1$ for every positive integer $n$ if and only if $k=1$.
[Hint]: Show that every prime must divide $2^n+3^n+6^n-1$ for some natural n. Consider cases when $p=2,3$ and $ > 3$. When $p > 3$, you may want to particularly look at $6(2^n+3^n+6^n-1)$, and to choose a value of $n$ which will allow you to use $\mathrm{F}\ell\mathrm{T}$.
Remark. 这里 $\mathrm{F}\ell\mathrm{T}$ 指 Fermat Little Theorem.