Find all values of positive integers $x,y,z$ so that $4^x+4^y+4^z$ is a perfect square I have to solve the equation
$$4^x+4^y+4^z=k^2$$
I posted my solution but I don't know if there are other proofs. How can I demonstrate that this expression is a perfect square?
Thanks in advance.
 A: We seek solutions to the Diophantine equation:
$$4^x + 4^y + 4^z = k^2$$ where $x, y, z$ and $k$ are integers. Let us assume that $x$ is the smallest of the set ($x, y, z$). Dividing both sides of the equation by $4^x$ (which is a perfect square) and rearranging terms yields:
$$4^u + 4^v = m^2 -1 = (m - 1)(m + 1)$$
Now the LHS is odd, only if $u = 0$ and $v > 0$ (or vice versa). But then the equation has no solutions. We may therefore assume that the LHS is even. This means that $m$ must be odd. Substituting $m = 2p + 1$ yields:
$$4^u + 4^v = 4p(p+1)$$
This equation is easily solved: $p$ must be of the form $4^q$, with $q = 0, 1, 2, 3,...$. The corresponding pairs ($u, v$) are found to be ($q+1, 2q+1$). The general solution for sets $(x, y, z)$ is therefore: 
$$(x, y, z) = (n, n + q + 1, n + 2q + 1)$$ where $n$ and $q$ are integers equal to or larger than $0$. If we focus on $n = 0$, the first five values for $k^2$ are found to be: $9 = 3^2, 81 = 9^2, 1089 = 33^2, 16641 = 129^2, 263169 = 513^2$.     
A: The equation that I have to solve is:
$$4^x+4^y+4^z=k^2$$
Manipulating the equation it becomes: $$4^x(4^{|y-x|}+4^{|z-x|}+1)=k^2$$
Now $4^x$ is always a perfect square, therefore we have to find value of $x,y, z$ so that $4^{|y-x|}+4^{|z-x|}+1$ is a perfect square.
We put $u=|y-x|$ and $v=|z-x|$ and the equation becomes $$4^u+4^v+1=m^2(k^2|m^2)$$
and
$$4(4^{u-1}+4^{v-1})=(m+1)(m-1)$$
Now we analyse three systems of equations:
$1)$
$$
\left\{ 
\begin{array}{c} 4=m-1\\  4^{u-1}+4^{v-1}=m+1\end{array}
\right.$$
$2)$ $$
\left\{ 
\begin{array}{c} 4=m+1\\ 4^{u-1}+4^{v-1}=m-1 \end{array}
\right.$$
$3)$ 
$$
\left\{ 
\begin{array}{c} 4=(m-1)(m+1)\\ 
4^{u-1}+4^{v-1}=1 \end{array}
\right.$$
From the first system, if $4=m-1$ then $m=5$, but $4^u+4^v+1=5$ hasn't solutions. 
From the second system, if $4=m+1$ then $m=3$, therefore $4^u+4^v+1=9$ has as only solutions $u=1$ and $v=1$. 
From the third system, if $4=(m+1)(m-1)$ then $m=\sqrt 5$ that isn't a integer.
If $u=1=|y-x|$ and $v=1=|z-x|$ the solutions are $z=y=x\pm 1$.
