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Solutions of the Diophantine equation

$a10^n+(a+1) = (2^{m+1}-1)*2^{m+1}$

are

12=3*4,

56=7*8,

67100672=8191*8192.

Are there more solutions/examples like that or a generalization of the equation? Does it describe a natural phenomenon?

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  • $\begingroup$ Note also that 672 = 12*56. $\endgroup$ – DVD May 13 '13 at 10:24
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    $\begingroup$ I think OP forgot a condition: $a<10^n$ $\endgroup$ – Next May 13 '13 at 10:35
  • $\begingroup$ There is no more solution if $a<10^n$ and $n<3000$. $\endgroup$ – Next May 13 '13 at 12:11
  • $\begingroup$ @Hecke Did you write code or you did estimates? $\endgroup$ – DVD May 13 '13 at 12:27
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$1.$ Denote $t=2^{m+1},N=10^n.$ If $a<N$ then $1\leq a=\dfrac{t^2-t-1}{N+1}<N,$ hence $$N+1\leq t^2-t-1<N(N+1),$$with some discussion we will get $$\sqrt{N}\leq t<N,10^{\frac{n}{2}}<2^{m+1}<10^n,$$ $$\frac{n}{2}\log_2{10}<m+1<n\log_2{10}$$ So for every given $n$,we can try all the $m$ one-by-one.I did this for $0<n<3000$,but no other solution was found.

$2.$ If $a<10^n$ is not necessary, we can use the following method for some little $n$.

Denote $t=2^{m+1},d=10^n+1,$ then $t^2-t-1\equiv0 \pmod d,$$$(2t-1)^2\equiv5 \pmod d\tag1$$ hence if $(1)$ has no solution,then $n$ is not the solution for the original problem.

If $(1)$ has some solutions, such as $t$,then $$2^{m+1}\equiv t \pmod d \tag2$$ If $(2)$ has solutions for $m$,then we get a solution for the original problem, and $m^{'}=m+k\phi(d)$ is a solution,too.

For example:

$n=5,m+1=13,a=671$

$n=2,m+1=64,a=3369132345751865974702256072852065939.$

$n=2,m+1=86,a=59270403034726518346161316806283592011628044701331.$

$n=5,m+1=4532,a=34346\cdots 204319.$

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