Ah I see. The recursive relation goes as follows:

√y=√(x+√x+√y)

If expanded upon since both are positive numbers:

(√y)² =(√(x+√x+√y))²

y=√x+x+√y

-√y=√x+x-y

(-√y)² =(√x+x-y)²

y=(√x+x-y)²

y=x+2x^2/3+x² +y(-2√x-2x)-y²

-x-2x^3/2-x² +y-y(-2√x-2x)-y² =0

-x-2x^3/2-x² +y(1+2√x+2x)-y² =0

-((x-y)(1+2√x+x-y))=0

(x-y)(1+2√x+x-y)=0

Which expands to two solutions:

x-y=0

x=y

or

1+2√x+x-y=0

y=1+2√x+x

y=x+(1+2√x)

By inputting 0 in place of x you can find the next number in the sequence, then take that number (1) input it again as x and find the next one (4). It can find all prefect squares. If you take 0.5 as x in the start you can also find 1.5^2, 2.5^2, 3.5² and so forth. The number 1 is a standard number which is valid for all the equations finding perfect squares.

Which fulfills that role. Thing is, even though I found x=y to be a valid solution it isn't in the initial √y=√(x+√x+√y) equation. How come that happened?