Powers of $19$ and $17$ differ by $2$

Question

Solve the equation $$19^x=17^y+2$$ over the positive integers.

This is strangely nonstandard (since working mod powers of $17$ or $19$ does not work). I think working in $\mathbb{Z}\left[\sqrt{2}\right]$ might be useful.

Answer 1

Got it, took a while.

We write $$ 19 (19^u - 1) = 17 ( 17^v -1) $$ We ASSUME $u$ and $v$ are nonzero, that is $u \geq 1$ and $v \geq 1$.

Steps... we need $19^u \equiv 1 \pmod {17}.$ Thus $$ 8 | u $$ We check, $19^8 - 1$ is divisible by $181,$ thus $19^u - 1$ is divisible by $181.$

We need $17^v \equiv 1 \pmod {181}.$ Thus $$ 36 | v $$ We check, $17^{36} - 1$ is divisible by $307,$ thus $17^v - 1$ is divisible by $307.$

We need $19^u \equiv 1 \pmod {307}.$ Thus $$ 51 | u $$ and especially $$ 17 | u $$ So far, we have $u$ divisible by both $8$ and $17$ so it is divisible by $136.$ We check, $19^{136} - 1$ is divisible by $17^2$ thus $19^u - 1$ is divisible by $17^2.$

However, the original way I wrote this was $$ 19 (19^u - 1) = 17 ( 17^v -1) $$ ASSUMING $u$ and $v$ are nonzero. Once $v \geq 1$ we see $17^v - 1 $ is NOT divisible by $17,$ so $17 ( 17^v -1) $ is NOT divisible by $17^2.$

This contradiction shows that the assumption is false, indeed $v=0$

P.S. we could have stuck with smaller exponents in places. For example we did not need to keep the exponent $36,$ actually $17^3 - 1$ is divisible by $307$ because $17^2 + 17 + 1 = 289 + 17 + 1 = 290 +17 = 307$

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