<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] $$2^a+2^b+2^c=2^c(2^{a-c}+2^{b-c}+1)$$
Since the second factor must be even, we see that $a>b=c$, and then it becomes
$$2^c(2^{a-c}+2)=2^{c+1}(2^{a-c-1}+1)$$
And now we see that $a=c+1$.
Now,
$$2^a+2^b+2^c=4\cdot2^c$$
so $c$ is even.
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