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Find all the nonnegative integers $a,b,c$ such that $1+2^a = 4 \cdot3^b + 5^c$.

I found this problem in an old number theory problem set. Using a computer, I found that the only solutions for $a,b,c \leq 30$ are $(2,0,0),(3,0,1),(4,1,1),(7,0,3),(12,5,5)$. I'm tempted to say that these are the only ones, but I haven't been able to prove it.

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Gabriel Romon
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  • This can be seen as an $S$-unit equation, so it follows that it must have at most finitely many solutions (the way it's given excludes all non-degenerate solutions other than $(2,0,0)$). There are some upper bounds for the number of solutions but they are quite big and the only bound I know for the size of the solutions is ineffective, so it can't be used to find them all by brute force. I'd be curious to know more, as well. – A.P. Dec 25 '17 at 23:37
  • There is an elementary approach, for fixed primes $p,q$ and fixed target $t,$ to bounding integers $m,n \geq 0$ in $p^m - q^n = t.$ I did not make it up, found on this site. I don't have anything for three primes. – Will Jagy Dec 26 '17 at 00:40
  • example, on the difficult side, but still elementary: http://math.stackexchange.com/questions/1946621/finding-solutions-to-the-diophantine-equation-7a-3b100/1946810#1946810 – Will Jagy Dec 26 '17 at 01:29
  • For $a,b,c\le 1000$ , no further solution exists. So, the list should be complete. – Peter Dec 26 '17 at 09:55
  • Where does that bound come from, @Peter? – A.P. Dec 26 '17 at 20:08
  • @A.P. Brute force with PARI/GP – Peter Dec 27 '17 at 18:38

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Observe that all the solutions appear to have $b \leq 5$. Treat the cases with $0 \leq b \leq 5$ as Will suggests (by local means or using lower bounds for linear forms in two logarithms), to reach the conclusion that the only solutions with $b < 6$ are the ones you have already found. Suppose that $b \geq 6$; note that both $2$ and $5$ are primitive roots mod $3^6$, having order $486$. We find precisely $243$ pairs of residue classes modulo $486$ for $(a,c)$ such that $$ 1+2^a \equiv 5^c \mod{3^6}. $$ Considering the original equation modulo various primes $p \equiv 1 \mod{486}$, we find that only the pair $a \equiv 248 \mod{486}$ and $b \equiv 427 \mod{486}$ is compatible with $1+2^a=4 \cdot 3^b+5^c$ modulo both $p=487$ and $p=1459$. Considering the equation modulo $p=17497$ then provides a contradiction.

Mike Bennett
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