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It is clear that $a>b \ge 1$ . Looking mod $b$ gives $b \mid a^{n-1} (a-1)$ and since $\gcd(a, b)=1$ we get $b|a-1$ and $a=bc+1$ for a positive integer $c$ . Looking mod $a$ gives $a \mid b^2+1$ . Thus $a \mid b^2+1-a \ge 0$ . Or $a \mid b^2+1-(bc+1)=b(b-c)$ and $a \mid b-c$ . But for $b-c>0$ we get $b+1 \le a \le b-c$ , which is a contradiction. Hence $b=c$ and $a=b^2+1$ .

The equation becomes $(b^2+1)^n-b^{n+1}=(b^2+b+1)^{n-1}$ Suppose that $p$ is a prime divisor of $b^2+b+1$ . Looking mod $p$ gives $‎(-b)^n ‎‎\stackrel{‎p‎}{\equiv} b^{n+1}$ or $‎b ‎‎\stackrel{‎p‎}{\equiv} (-1)^n$ . For $n$ being odd, it follows that $‎b+1 ‎‎\stackrel{‎p‎}{\equiv} 0$ , which is impossible since $\gcd(b+1, b^2+b+1)=1$ . Thus $n$ is even and $p| b-1$ and $p \mid b^2+b+1-b(b-1)=2b+1 \\ p \mid (2b+1)-2(b-1)=3$ So $p=3$ is the only prime divisor of $b^2+b+1$ and $b^2+b+1=3^k$ for a positive integer $k$ .

We have $(2b+1)^2=4\times 3^k-3$ . LHS of this equation is divisible by $9$ , but for $k>1$ RHS is not divisible by $9$ . Thus $k=1$ . It follows that $b=1$ and $a=2$ . Now, the original equation is $2^n-1=3^{n-1}$ . Its immediate that $n=2$ .

Thus the only solution is $(a, b, n)=(2, 1, 2)$ and answer is $\boxed{5}$ .