$ 0=1 $ ? Where is the mistake? I just found this formula although it can be easily derived.
Let $ n $ be any integer then,
$$n=\sqrt{n^2-n+\sqrt{n^2-n+.....}}$$
So if I plug in $ 0 $ in this equation I get,
$$0=\sqrt{0-0+\sqrt{0-0+....}}$$

$$0=\sqrt{0+\sqrt{0+\sqrt{0+....}}}$$—————->1

But if I plug in $1$ in the equation I get,
$$1=\sqrt{1^{2}-1+\sqrt{1^{2}-1+.....}}$$

$$1=\sqrt{0+\sqrt{0+\sqrt{0+....}}}$$,—————->2

Which gives me that 0=1.Where is the mistake?
Edit:
Equation 1 can also be derived by the following method
$$1=\sqrt{0+\sqrt{1}}$$
$$1=\sqrt{0+\sqrt{0+\sqrt{1}}}$$
.
.
.
$$1=\sqrt{0+\sqrt{0+\sqrt{0+....}}}$$
Ok so let me tell you how I derived it.
I came across this particular expression,
$$x=\sqrt{1+\sqrt{1+\sqrt{1+.....}}}$$ which actually gives me the golden ratio,
So I chose n to be an integer and let x be the value of the following expression,
$$x=\sqrt{n+\sqrt{n+\sqrt{n+...}}}$$.
After solving for x you get it’s value to be!
$$x=\frac{1+\sqrt{1+4n}}{2}$$
(I took a positive sign since x is greater than or equal to 0)
So x can be an integer whenever n=0,2,6,10,20.....
When n=2,x=2(since x>0) and we get,
$$2=\sqrt{2+\sqrt{2+......}}$$
Similarly for n=6,x=3,
$$3=\sqrt{6+\sqrt{6+......}}$$
So x is an integer whenever $$n=k^{2}-k$$ where k is a non-negative integer.Substituting $$n=k^{2}-k$$ you get x=k.
After all this I get,

$$n=\sqrt{n^2-n+\sqrt{n^2-n+.....}}$$


Assuming convergence:
$$a=\sqrt{n^2-n+\sqrt{n^2-n+.....}}$$
$$a=\sqrt{n^2-n+a}$$
$$a^2=n^2-n+a$$
$$a^2-a=n^2-n$$
$$a(a-1)=n(n-1)$$
For the case $n=1$:
$$a(a-1)=0$$
Notice that $a$ has roots $1$ and $0$. Why did you assume $a$ to be one of them, $1$? All $a$ that satisfies $a=\sqrt{n^2-n+a}$ and is positive is correct.
