<s>[INST] Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$? This question arose on the code golf StackExchange:

Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$?

In other terms: is there an integer that looks like $\color{blue}{1000...001000...}$ in both binary and decimal?
I feel like there probably isn't, but I can't think of a simple counterargument.
(A computer search by one of the commenters suggests there is no such integer up to $10^{100000}$.)
 [/INST] NOT AN ANSWER. Commentary too long to fit in a Comment

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*The equation: $2^a+2^b=10^c+10^d;\ 0\le a<b,\ 0\le c<d$. If $b>a,\ d>c$ were not already stated, it is provable that $a\ne b,\ c\ne d$ from which we could validly assume WLOG $b>a,\ d>c$.


*Although $a=0$ and $c=0$ are contemplated by the question as stated, neither $a$ nor $c$ can actually be $0$. If only one of them is, then we have an odd sum equal to an even sum. If both of them are, then by subtracting $1$ from each side we obtain $2^b=10^d$ which is impossible.


*$b-a=4k-2$. This follows from $2^a+2^b\equiv 0 \bmod 5$. Positive integer powers of $2$ modulo $5$ cycle in order through $2,4,3,1$. Sums of powers of $2$ that are $\equiv 0 \bmod 5$ arise from $2^{4m+1}+2^{4n+3}\text{ or }2^{4m}+2^{4n+2}$, and the differences of the exponents have the stated form.


*$a=c$. The original equation can be rearranged to $2^a(2^{b-a}+1)=2^c5^c(10^{d-c}+1)$. Plainly, there are $a$ factors of $2$ on the LHS and $c$ factors of $2$ on the RHS, so $a=c$.


*$a$ is even. From point 4 we obtain $2^a+2^b=10^a+10^d \Rightarrow 1+2^{b-a}=5^a(1+10^{d-a})$. We look at this equation $\bmod 3$. $1+2^{b-a}\not \equiv 1 \bmod 3$, and $(1+10^{d-a})\equiv 2 \bmod 3$. This requires $5^a \not \equiv 2 \bmod 3$ which in turn requires $a$ is even. This impacts point 3 by restricting the exponents to $(a,b)=(4m,4n+2)$ up to order. So $b$ is also even.


*$d$ is even. From points 4 and 5, $c$ is even. $10\equiv -1 \bmod 11$ so $10^c+10^d \equiv 1+(-1)^d \equiv 0,2 \bmod 11$; $0$ if $d$ is odd and $2$ if $d$ is even. $2^{4m}+2^{4n+2}\not \equiv 0 \bmod 11$. There are instances where $2^{4m}+2^{4n+2} \equiv 2 \bmod 11$. So if there are solutions to the equation, $d$ is even.
This is as far as I have gotten; I will update if I get further.
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