In thinking about a MathOverflow question pertaining to numbers whose decimal and binary digit sums are equal, I found myself asking:

Are there any solutions in non-negative integers $(a,b,c,d)$ to the equation

$$2^a + 2^b = 10^c + 10^d$$

aside from the trivial solution $(0,0,0,0)$ and the nontrivial solutions $(2,4,1,1)$ and $(4,2,1,1)$?

(This is, in effect, the case $s=2$ to the other MO question regarding which numbers $s$ can be a simultaneous binary and decimal digit sum and if so, how often. It seems likely that 20 is the only number whose digit sums are both 2, but that's basically what I'm asking here.)

In searching for other solutions, we may as well assume, for the sake of simplicity, that $a\le b$ and $c\le d$. It's quickly clear that we can restrict to looking for *positive* solutions, and we can also dismiss the possibility that $a=b$. That is, we can assume $0\lt a\lt b$ and $0\lt c \le d$.

The possibility that $c=d\ne 1$ is ruled out by Mihăilescu's proof of Catalan's conjecture: If $2^a + 2^b = 2\cdot10^c = 2^{c+1}\cdot5^c$ (with $a\lt b$) we necessarily have $a=c+1$, leaving the equation $1 = 5^c - 2^n$, where $n=b-a$, whose only solution is $c=1$, $n=2$. (Aside: We don't actually need to invoke Catalan's conjecture. See below.)

If now we restrict to $c\lt d$, it's immediately clear that we must have $a=c$. After writing $n=b-a$ (as before) and $m=d-c$ and doing a little factoring, the problem reduces to what I take to be the core question:

Does the equation

$$1+2^n = 5^a(1+10^m)$$

have any solutions in positive integers $(a,m,n)$?

This is as far as I've gotten in any meaningful sense. The question is obviously related to the Cunningham project. It's easy to see that $a>0$ implies $n\equiv2\mod4$, so the aurifeuillian factorization

$$2^{4k+2}+1 = (2^{2k+1}-2^{k+1}+1)(2^{2k+1}+2^{k+1}+1)$$

may or may not help. (It *does* help avoid the use of invocation of Catalan's conjecture above: Only one aurifeuillian factor is divisible by 5.) It's also easy to see that

$$n\log2 + \log(1+1/2^n) = a\log5 + m\log10 + \log(1+1/10^m)$$

implies

$$(a+m)\log5 \approx (n-m)\log2$$

if $m$ and $n$ are large. Finally, writing $5^a = (1+4)^a = 1+4a+16{a\choose2}+\cdots$, it may be possible to say something useful about the relationship of $a$ to $n$, but I don't see what.