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?
 A: If we write the numbers in binary system we get
$$1\underbrace{0\ldots0}_a+1\underbrace{0\ldots0}_b+1\underbrace{0\ldots0}_c=1\underbrace{0\ldots0}_{2d}$$
This is possible only if $b=c$ to get $1+1=10$ and $a=b+1$ to make the $1$ dissapear with the carry.
In addition, $a$ must be odd.
A: 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 $b=c=2d-2$, $a=2d-1$.
A: I'd write this as
$$\frac1{2^r}+\frac1{2^s}+\frac1{2^t}=1$$
where $r=2d-a$, $s=2d-b$ and $t=2d-c$ are integers. Each of $r$, $s$ and $t$
must be positive, and they can't be all $\ge2$, since then the LHS is $\le 3/4$.
So one of them, say $r$, equals $1$. In that case
$$\frac1{2^s}+\frac1{2^t}=\frac12.$$
The only possibility here is $s=t=2$. So the solutions for $(r,s,t)$ are
$(1,2,2)$, $(2,1,2)$, $(2,2,1)$.
What about $(a,b,c)$? If $(r,s,t)=(1,2,2)$ then $(a,b,c)=(c+1,c,c)$
and $2^a+2^b+2^c=2^{c+2}$. Then $c=2d-2$ must be even, and $(a,b,c)
=(2d-1,2d-2,2d-2)$ etc.
A: WLOG $a\ge b\ge c$. If $b\neq c$ then consider $\pmod {2^{c+1}}$ we get a contradiction (why?). Then observe $a=b+1$ by considering $\pmod{2^{b+2}}$. Then $a=b+1=c+1$, so $2^b=4^d\implies b$ is even. Hence the solution you found are the only solutions.
A: $$2^a+2^b+2^c=2^c(2^{a-c}+2^{b-c}+1)$$
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.
A: 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.
