I observed that if we have
a) $ \gcd (a, \phi (a))=\phi(a)$ and $ \gcd (b, \phi (b))=\phi(b)$ and $ \gcd (c, \phi (c))=\phi(c)$
b) and $\phi(a^2)+\phi(b^2)=\phi(c^2)$
that then we also have:
$$(\phi(a))^2+(\phi(b))^2=(\phi(c))^2$$
Is there a finite or infinite number of triples $(a,b,c)$ of natural numbers for which a) and b) are simultaneously satisfied? How many triples can you find?
Remark: The condition $ \gcd (n, \phi (n))=\phi(n)$ implies that $\phi(n)\mid n$, so that $n=2^k3^l$, see here:
How do you find all $n$ such that $\phi(n)|n$
Then $\phi(n)=\phi(2^k)\phi(3^l)=2^{k-1}\cdot 2\cdot 3^{l-1}$. Now write $a,b,c$ as powers of $2$ and $3$ and compute.