<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] We have
$$
\eqalign{
  & \left( {a,b,c,d} \right)\quad  \Rightarrow \quad 2^{\,a}  + 2^{\,b}  + 2^{\,c}  = 4^{\,d} \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad 2^{\,2n} \left( {2^{\,a}  + 2^{\,b}  + 2^{\,c} } \right) = 2^{\,2n} 4^{\,d} \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad 2^{\,a + 2n}  + 2^{\,b + 2n}  + 2^{\,c + 2n}  = 4^{\,d + n} \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {a + 2n,b + 2n,c + 2n,d + n} \right) \cr} 
$$
So the problem shifts to find the minimal quadruples.
In binary notation it is clear how it shall be
$$
\eqalign{
  & 2^{\,a} \quad \quad \quad \quad  = 0,0, \cdots ,1,0,0  \cr 
  & 2^{\,b} \quad \quad \quad \quad  = 0,0, \cdots ,1,0,0  \cr 
  & 2^{\,a}  + 2^{\,b} \quad \quad  = 0,0, \cdots ,0,1,0  \cr 
  & 2^{\,c} \quad \quad \quad \quad  = 0,0, \cdots ,0,1,0  \cr 
  & 2^{\,a}  + 2^{\,b}  + 2^{\,c}  = 0,0, \cdots ,0,0,1  \cr 
  & 4^{\,d}  = 2^{\,2d} \quad \;\; = \underbrace {0,0, \cdots ,0,0,1}_{0,1, \cdots ,\;2d - 2,\;2d - 1,\;2d} \cr} 
$$
So
 - $2^k$ has only one $1$, and same has $4^d$;
 - in summing the $2$'s, to reach to have only one $1$, we cannot leave any $1$ behind;
 - so the (only) one of $2^b$ must be just below that of $2^a$
 - ... and we must end on bit $2d$.
The conclusion is clear.
</s>