<s>[INST] Find $a,b,c,d$ such that $2^a + 2^b + 2^c = 4^d$ Let $a,b,c,d$ be whole numbers that satisfy 
$$2^a + 2^b + 2^c = 4^d$$
What values of $(a,b,c,d)$ would make this equation true?

Here is my work so far.
Without loss of generality, assume $a\ge b\ge c$. Then one trivial solution by inspection is $(1,0,0,1)$. Playing around, I also found a solution at $(3,2,2,2)$. Then I checked $a=5,b=4,c=4$ and found that it also worked.
It seems that there is a family of solutions at $(2n-1,2n-2,2n-2,n)$. I can prove this easily:
\begin{align}
LHS&=2^{2n-1}+2^{2n-2}+2^{2n-2}\\
&=2^{2n-1}+2^{2n-1}\\
&=2^{2n}\\
&=4^n\\
&=RHS
\end{align}
Is this the only solution? If it is, how do I go about proving it?
 [/INST] Obviously, $d>0$. WLOG, assume that $a\ge b\ge c$
$2^c(2^{a-c}+2^{b-c}+1)=2^{2d}$
$2^{a-c}+2^{b-c}+1\ge 3$. Therefore, $b-c$ must be $0$ as otherwise $2^{a-c}+2^{b-c}+1$ will be an odd number larger than $1$.
$2^{a-c}+2^{b-c}+1=2^{a-c}+2=2(2^{a-c-1}+1)$ and hence $2^{a-c-1}+1$ must be even. $a-c=1$.
We have $2^c\times 4=2^{2d}$ and hence $b=c=2d-2$, $a=2d-1$.
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