how compute Phi(N) of RSA?

[BirdKing]

hello, I have a problem about RSA. Please help me!
IN RSA,
N=P*Q
Phi(N) = (P-1)*(Q-1)

When compute standard RSA, a necessary condition is D<Phi(N).
Phi(N) = (P-1)*(Q-1) = P*Q - (P+Q) + 1
according N = P*Q, P + Q ≈ 2*（N^½）
Phi(N) = N - 2*(N^½) + 1

The problem is how to compute N^½

[horse dream]

In RSA computation,
N=P*Q
Phi(N) = (P-1)*(Q-1)

When computing RSA STD, D<Phi(N)
Phi(N) = (P-1)*(Q-1) = P*Q - (P+Q) + 1

According to N = P*Q, P + Q ≈ 2*（N^?）
Phi(N) = N - 2*(N^?) + 1

How to compute N^?:
According to X2 = [X1 + (N/X1)]/2,
The first computation, X1 takes a radom number, compute X2.
The second computation, save X1： X3 = X1, give X2 value to X1: X1 = X2, compute X2 again.
Continue computing until (X1 >= X2 and X1 >= X3) or (X1 <= X2 and X1 <= X3). And now X1 is relatively close to N^?.

e.g.
N = 15
X1 = 7, ==>> X2 = 4
X3 = 7, X1 = 4, X2 = 3
X3 = 4, X1 = 3, X2 = 3
So the value taken finally is 3.

Another example：
N = 15
X1 = 2, ==>> X2 = 4
X3 = 2, X1 = 4, X2 = 3
So the value taken finally is 4.

[BirdKing]

Hello,
Thank you very much.
I solved the problem following your method.