Can be rational? Can $z=\sqrt{1-4y^x}$ be a rational number, where $x>2$ is an integer and $y$ is a rational number and $y>0$?
I have tried with $x=2$ and it has rational solutions, for example $y=12/25$ and then $z=7/25$. But I haven't found any rational solution with $x>2$
 A: If $x$ is an even integer, then $x=2k$ for some integer $k$, $z^2+(2y^k)^2=1$, for some $a$, $b$ and $c$ put $y=b/c$ and $z=a/c^k$, so, $a^2+(2b^k)^2=(c^k)^2$ this is pythagorean  with solutions $a=u^2-v^2$, $c^k=u^2+v^2$ and $b^k=uv$. From the last equation, $u$ and $v$ must be some power of $k$ (since they are coprime), say, $t^k$ and $s^k$ respectively. Hence, the second to last equation becomes $(c)^k=(t^2)^k+(s^2)^k$ and has no solution for $k>2$ by Fermat's Last Theorem, hence $k=0, 1, 2$, if $k=0$ the original equation has no solution since $z$ will be complex valued, if $k=1$ then $x=2$ but this contradicts the stated condition for $x$. Hence $x=4$ is the only candidate for examination. The original equation becomes; $z^2+4y^4=1$ which a little analysis shows has no rational solution by transforming it to one of Fermat's equations in his proof of FLT for the exponent four $n=4$. Hence, if $x$ is even, the stated equation has no solution
A: Too long for a comment:
$\sqrt{1-4y^x}=z\quad=>\quad1-4y^x=z^2\iff1-4~\bigg(\dfrac mn\bigg)^x=\bigg(\dfrac ab\bigg)^2,\quad\begin{cases}\gcd\big(~a~,b\big)=1.\\\gcd\big(m,n\big)=1.\end{cases}$
$\iff\underbrace{n^x\overbrace{(b^2-a^2)}^{b~>~a}=\overbrace{4b^2m^x}^{>~0}}_{\gcd(a,b){\large=}1,\quad\gcd(m,n){\large=}1.}~=>~n^x$ can only be given by $4b^2$, and $b^2-a^2$ can only come from 
$4m^x$. Now, $4$ can be broken up into a product of two factors in only two ways: as $1\cdot4$, or as $2\cdot2$, 
giving us the three possible cases $~\begin{cases}~n^x=4b^2\\m^x=b^2-a^2\end{cases}\quad,~\begin{cases}~~~n^x=2b^2\\2m^x=b^2-a^2\end{cases}\quad,~\begin{cases}~~~n^x=b^2\\4m^x=b^2-a^2\end{cases}$
