On $\phi(n) \sigma(n)$ being a square. 
I can't understand underlined statements: 
The green one: For p an odd number, both p-1 and p+1 are even so all prime factors in $\prod (p-1)(p+1)$ must be belong to the set B, so the prime factors of $m_P$. So $\phi(n_P) \sigma(n_P) = mt_P^2$ holds for any collection of primes selected from set A. Where does $2^{\pi(y)-\pi((y+1)/2)}$ come from?
The red one: Why $m$ is same for $n_P$ and $n_Q$? 
The blue one: How it is injective? How the identity element is empty? How $\Delta$ and $A$ make a group?
 A: Orange/red: This is how the set $\mathcal{F}\subseteq P(\mathcal{A})$ is defined: every $\mathcal{P}\in\mathcal{F}$ has the same $m$.
Green: $2^{\pi(y)}$ and $2^{\pi((y+1)/2)}$ are the cardinality of $P(\mathcal{A})$ and $P(\mathcal{B})$, as stated in the second/third line in the paragraph.  Since $m$ is a product of elements from $\mathcal{B}$, there are at most $\lvert P(\mathcal{B})\rvert$ choices of $m$, but there are $\lvert P(\mathcal{A})\rvert$ choice of $\mathcal{P}$ so there is some $m$ which is hit by at least $\dfrac{\lvert P(\mathcal{A})\rvert}{\lvert P(\mathcal{B})\rvert}$ elements of $P(\mathcal{A})$.
Blue: You can recover $\mathcal{P}$ from $\mathcal{P}\triangle\mathcal{Q}$ by symmetric difference again with $\mathcal{Q}$.  Alternatively, you can view $2$ as the group of order $2$ and the power set $P(\mathcal{F})=2^{\mathcal{F}}$ as the direct product of $\lvert\mathcal{F}\rvert$ copies of $2$, then the group operation coincides with symmetric difference.
A: COMMENT.-Some questions that would imply   $\textbf{Proposition 8.7}$
Taking $n=pq$ with $p,q$ distinct primes one has $\phi(n)=(p-1)(q-1)$ and $\sigma (n)=1+p+q+pq=(p+1)(q+1)$ so 
$$\phi(n)\sigma (n)=(p^2-1)(q^2-1)$$ 
For the primes $2$ and $3$ one has
 $$\phi(2p)\sigma (2p)=3(p^2-1)\\\phi(3p)\sigma (3p)=2^3(p^2-1)$$ Then in order to have   $\phi(2p)\sigma (2p)=\square$ we need primes $p$ such that $p^2-1=3x^2$. The two first are  $7$ and $97$.
Similarly for $\phi(3p)\sigma (3p)=2^3(p^2-1)$ we need primes $p$ such that $p^2-1=2x^2$. The two first are  $17$ and $577$.
►Are there infinitely many primes $p$ such that $p^2-1=3x^2$? or $p^2-1=2x^2$?
This question can be extended fixing all prime instead of $2$ and $3$ above but sometimes the coefficient $ a $ in $ ax ^ 2 $ can be large (for $ n = 61p $ you have $61^2-1 =2^3\cdot3\cdot11\cdot17$ for which $a=2\cdot3\cdot11\cdot17 = 930$ and surely  there are many more possible larger values. However there are many other primes like $19$ for which $\phi(19p)\sigma (19p)=360(p^2-1)$ so we need primes such that $p^2-1=10x^2$. Note that a single solution in an infinity of these cases would also prove the proposition. $$*****$$
We can verify that for $n=2p,7p$ and $97p$ we need primes such that $p^2-1=3x^2$ to have $\phi(2p)\sigma (2p),\phi(7p)\sigma (7p)$ and $\phi(97p)\sigma (97p)$ squares. In other words, a solution for one of three is also solution for the other two. Analogously for $n =3p$ and $n=17p$ we need to solve the same equation $p ^ 2-1 = 2x^2$.
►Are there infinitely many primes $p_i$ such that $p_i^2-1=by^2$ where the square-free part $b$ is the same for each $p_i$? 
In this case just one solution for any of the $\phi(p_i\cdot p)\sigma (p_i\cdot p)=\square$ would prove the proposition.
