They Are All Related To Each Other!

Relevant wiki: System of Equations - Problem Solving - Intermediate

I shall first prove that there are no solutions when any 2 of these variables are not equal. WLOG, suppose $x>y$ . Then $\sqrt{4x-1}>\sqrt{4y-1}\Rightarrow y+z>x+z \Rightarrow y>x$ . Contradiction! Hence all of the variables are equal, so $2x=\sqrt{4x-1}\Rightarrow 4x^2-4x+1=0\Rightarrow (2x-1)^2=0\Rightarrow x=y=z=1/2$

$As,$ $x=y=z$

$Squaring$ $every$ $equation$

$(4x^2 = 4x-1)*3$

$then$ $12x^2-12+3=0$

$x=1/2$

$S0,$ $only$ $one$ $ordered$ $triples$ $stands$

(x + y)^2 = 4z - 1,
(x + y)^2 - 2(x + y) + 1 = 4z - 2(x + y),
(x + y - 1)^2 = 4z - 2(x + y),.......I,
Likewise, the other two equations will give, (y + z - 1)^2 = 4x - 2(y + z),.......II, and,
(z + x - 1)^2 = 4y - 2(z + x),........III,
Adding I, II and III, we have,
(x+y-1)^2+(y+z-1)^2+(z+x-1)^2 = 0,
This is possible only when each of the term on LHS is zero. Therefore,
x + y = 1, y + z = 1, and, z + x = 1.
Adding these equations give,
x + y + z = 3/2, and, therefore, on solving, we have,
x = y = z = 1/2.

As $x+y = \sqrt{4z-1}, y+z=\sqrt{4x-1}, x+z = \sqrt{4y-1}$

Squaring on both the sides, $(x+y)^2 = 4z-1.....1$

$(y+z)^2=4x-1, ....2$

$(x+z)^2 = 4y-1......3$

Subtracting $1$ and $2$ , $2$ and $3$ , $3$ and $4$

$(x-z)(x+2y+z) = 4(z-x)$

$(x-y)(x+2z+y) = 4(z-x)$

$(y-z)(y+2x+z) = 4(z-y)$

So we will get, $x+2y+z = x+2z+y=y+2x+z = -4$

so, x= y=z. Now on substituting in the main equation,

$2x = \sqrt{4x-1}$ $4x^2 = 4x-1$

$x=\dfrac{1}{2}$ . Thus we have only one solution, $(\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2})$

Since we are told x=y=z, all equations are the same

We are left with

$x+x = \sqrt{4x - 1} >$

$2x = \sqrt{4x - 1} >$

$4x^2 = 4x - 1 >$

$4x^2 - 4x +1 = 0 >$

$(2x-1)^2=0 >$

$x=.5$

There is only 1 solution to this equation