<s>[INST] 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.
 [/INST] 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 $z=y=x\pm 1$.
</s>