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I want to find the solutions $(a,b)\in\mathbb{Z}^+\times\mathbb{Z}^+$ of $7^a+2=3^b$. One such solution is $(a,b)=(1,2)$.

Looking modulo $4$, we have $(-1)^a+2\equiv(-1)^b$, so $a$ and $b$ are of different parity.

Modulo $3$: $1+2\equiv 0$, which is always true.

Modulo $7$: $2\equiv 3^b$, so $b\equiv 2\pmod 6$. This means $a$ is odd.

Dexter
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  • Doing arithmetic modulo $;9;$ one gets that it must be $;a=1\pmod 3;$, otherwise the left side isn't zero modulo $;9;$ , whereas the right one obviously is. – Timbuc Nov 05 '14 at 19:32
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    This is the German 2010 VAIMO 1 problem 3. – Bart Michels Nov 06 '14 at 16:10
  • I rewrite the equation $7^a=3^b-2$ and calculate $a=log((3^b-2)/log(7)$ and calculate the discrepancy $d=3^b-floor ((7^( log((3^b-2)+2)/log(7))$ which is a measure of the distance from $3^b$ to the nearest $7^a+2$. I note that for b=2,3,4,...... the discrepancy is increasing but there seems to be an anomaly for (at least some) values of b near $b=2^n$ such as b=9,32,62,125,248,512 (but also others like b=39 and 48) where the value is not increasing, indicating a relation to the related equation $7^a+2=(3^2)^b=9^b$. I realize this is a little beside the point. – Mikael Jensen Nov 07 '14 at 22:59
  • Your question is similar to a question what I have posted here: http://math.stackexchange.com/questions/508714/solving-the-equation-x2-7y-2 . But these two equations have something different. – Travis Wang Nov 10 '14 at 08:34
  • I've written a loong answer at the current *duplicate-question*, see https://math.stackexchange.com/a/2272949/1714 . Perhaps it should be transferred to this earlier question? – Gottfried Helms May 09 '17 at 12:06
  • Searching using Approach0 also returns this question. – Martin Sleziak May 09 '17 at 12:57

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