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Find all positive integers $n$ and $m$ such that $(125\times2^n)-3^m=271$

I have thought about this question for a long time and I can't seem to solve it. I realize that $271$ is a prime and so I'm tried to factor the LHS but couldn't. I came to the conclusion that n is odd and m is equivalent to $2$ modulo $4$ by using modulo $3$ and $5$ respectively. Any help would be appreciated.

user587054
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  • Note; $(n,m)=(3,6)$ is one solution – J. W. Tanner May 17 '19 at 20:59
  • @J.W.Tanner I did realize that and I was able to solve for m and n when they were multiples of 3, but couldn't continue for other values. – user587054 May 17 '19 at 21:09
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    Note that with $p=n-3$ and $q=m-6$ (small solutions can be dealt with easily) we have $1000\cdot 2^p-729\cdot 3^q=1000-729$ so that $1000\cdot (2^p-1)=729(3^q-1)$ - there are prospects here, and modulo $1000$ and $729$ you get a sparser set of possibilities than modulo $3$ and $5$. – Mark Bennet May 17 '19 at 22:11
  • @MarkBennet sure but that still doesn't look at all fun to try to solve – Mike May 17 '19 at 22:35
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    @MarkBennet $q \equiv 0 \pmod {100},$ in particular $q \equiv 0 \pmod 4,$ then $3^q-1$ is divisible by $16,$ therefore $2^p - 1$ is even and $p=0$ – Will Jagy May 17 '19 at 22:38
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    @WillJagy you should write this out as an answer. This looks really nice – Mike May 17 '19 at 22:41

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I put a really difficult example of this method at Finding solutions to the diophantine equation $7^a=3^b+100$ including some links to easier examples and the place where I learned it; seems fair to give credit to the student who seems to have invented it, Exponential Diophantine equation $7^y + 2 = 3^x$

From Mark Bennet, we reach $$ 1000 (2^p - 1) = 729 (3^q - 1) $$ The argument goes back and forth, this one is quick: we know $$ 3^q \equiv 1 \pmod 5 \; , $$ so that $$ q \equiv 0 \pmod 4 \; . $$ However, then $3^4 - 1 = 81 - 1 = 80 = 16 \cdot 5,$ so $$ 3^q - 1 \equiv 0 \pmod {16}. $$ Now $$ 16 | 1000 (2^p-1) \; , $$ so $2^p - 1$ is even and $$ p=0$$

Will Jagy
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  • Nicely done. I was falling asleep at the screen and didn't follow i through ... – Mark Bennet May 18 '19 at 07:46
  • So a little more generally than the case in the link, if $ap^M-bq^N=r$ has the solution $ap^m-bq^n=r$ we can eliminate $r$ to obtain $ap^m(p^{M-m}-1)=bq^n(q^{N-n}-1)$ and this method can be used to investigate 'large' solutions. I am now wondering what might be known which would throw further light on this procedure and when it will work and if it ever fails. – Mark Bennet May 18 '19 at 08:10
  • @MarkBennet thanks. Gottfried came up with a different view of this. I'm not sure it ever fails, but I attempted one where the numbers got too big to continue with my computer/software. The way I do it is pretty random looking, go through all prime factors of some $p^d - 1,$ whee $d$ is large enough to make $p^d-1$ enormous. I have no idea how to predict the prime factor that should be picked for the next step... – Will Jagy May 18 '19 at 17:59