# A Diophantine equation and decimal digits

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?

• Note also that 672 = 12*56. – DVD May 13 '13 at 10:24
• I think OP forgot a condition: $a<10^n$ – Next May 13 '13 at 10:35
• There is no more solution if $a<10^n$ and $n<3000$. – Next May 13 '13 at 12:11
• @Hecke Did you write code or you did estimates? – DVD May 13 '13 at 12:27

$1.$ 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=671n=2,m+1=64,a=3369132345751865974702256072852065939.n=2,m+1=86,a=59270403034726518346161316806283592011628044701331.n=5,m+1=4532,a=34346\cdots 204319.\$