Infinite nested radicals

$\displaystyle y = \frac{1 + \sqrt{4z - 3}}{2}$

4z - 3 should be perfect square and the least possible value of z is $\boxed {3}$ When y is natural number

The given equation may be rewritten as:- $\sqrt{z+\sqrt{z-y}}=y$ Squaring both sides:- $\implies z+\sqrt{z-y}=y^{2}$ $\implies \sqrt{z-y} = y^{2}-z$ Squaring both sides again:- $\implies z-y = y^{4}-2y^{2}z+z^{2}$ Rearranging the above terms:- $\implies z^{2}-z(1+2y^{2}) + y + y^{4} = 0$ Solving by quadratic formula:- $\implies z=\dfrac{1+2y^{2}\pm \sqrt{1+4y^{2}+4y^{4}-4(y+y^{4})}}{2}$ Solving It we get:- $z=y+y^{2} or z=y^{2} - y +1$ We have to find the values of z for which y is a natural number.

Put $y =0$ in both the above equations, we get a $z=0$ and $1$ . Which are not allowed.

Put $y=1$ in both the above equations, we get $z=2$ and $1$ .

If we put $z=2$ in the given equation, we will find inconsistency. So we will neglect it.

Put $z=2$ in both the equations. We get $z=6, 3$ .

Put $z=3$ in the given equation. We get the equations are consistent.

Hence $z=3$ is the smallest values for which $y$ is a natural number.

First of all, observe that

$y^{2} = x + \sqrt{x-y} \;\;\; \Rightarrow \;\;\; y^{4} - 2y^{2}x + y + x^{2} - x = 0$

Now, a little theorem says that the only possible rational irreductible roots $\frac{a}{b}$ of a polynomial equation with integer coefficients $c_{n}x^{n} + \cdots + c_{1}x + c_{0} = 0$ are those in which $a$ divides $c_{0}$ and $b$ divides $c_{n}$ .

Therefore, for our polynomial in $y$ , the only possible roots are $y = x$ and $y = x - 1$ .

Testing for $y = x$ gives us $x^{4} - 2x^{3} + x^{2} = x^{2}(x^{2}-2x+1) = 0$ with $x \neq 0$ which has the only root $x = 1$ . But this is not the answer, according to the problem. Therefore, $y \neq x$ .

Testing for $y = x - 1$ we'll get $y^{2} = x + \sqrt{x-y} \;\;\; \Rightarrow \;\;\; x(x-3) = 0$ with $x \neq 0$ . Therefore $x = 3$ . Then

$x = 3$

Since y = √(z+√(z-√(z+√(z-√(z+⋯)) ) ) ) = √(z+√(z-y)) ,

y = √(z+√(z-y))

y^2 = z+ √(z-y)

y^2 - z = √(z-y)

Since √(z-y) should be a natural #, then, z-y >/= 0 ---> z >/= y ---->Least posible value of z: y+1. then,

y^2 = y+1+1 or y^2 = y+1-1

y^2 - y - 2 = 0 or y^2 - y = 0

y = {2,-1} or y = {0,1}

In y = {2,-1}, reject -1 since it's not a natural #. So, y = 2 is acceptable. then, z = 3

in y = {0,1}, reject 0 since it's not a natural #. so, y = 1 is acceptable. and as a result, z = 2

But, if you will recheck or substitute: (y,z) = (1,2) in y^2 = z+ √(z-y) , It will not satisfy the equation, only (y,z) = (2,3). So therefore, The smallest possible value of z is 3.

You should explain why the answer isn't 1.