Euler function of product of two prime squares

Question

Is there a shortcut for finding the Euler function of the product of two prime squares?

$\phi(a^2*b^2)$=?, where a and b are prime.

I am able to find $\phi(a*b)=(a-1)(b-1)$, but don't really know where to start next. Both a and b are very large, and $\phi$ will be used for encryption methods.

Answer 1

Well if $a \neq b$ and both are prime then $gcd(a^2,b^2) = 1$ and we can therefore break the $\phi$ function: $\phi(a^2 \cdot b^2) = \phi(a^2) \cdot \phi(b^2) = a(a-1) \cdot b(b-1)$.

See Wiki for the identity I used.

Answer 2

$\varphi(p_1^{n_1} \cdots p_m^{n_m}) = p_1^{n_1} \cdots p_m^{n_m}(1-1/p_1)\cdots(1-1/p_m)$.

Answer 3

I imagine that $*$ means the standard integers multiplication.

If yes, there is a standard formula for Euler's totient function. Namely $$\phi(n) = n\prod\limits_{p | n} \left(1-\frac{1}{p}\right).$$

Your $n=a^2b^2$. Hence $$\phi(a^2b^2) = ab(a-1)(b-1).$$