Note that $a$ and $n$ are relatively prime if and only if $n-a$
and $n$ are relatively prime.
Call such a pair $\{a,n-a\}$, where $a$ is relatively prime to $n$, a couple. The sum of the numbers in a couple is $n$.
If $n\gt 2$, there are $\frac{\varphi(n)}{2}$ couples. For every $a$ relatively prime to $n$ gets coupled with someone other than herself.
It follows that our sum is $n\cdot\frac{\varphi(n)}{2}$. If this is $240n$, then $\varphi(n)=480$.
Yes, the proof is correct.
Let $c=\Pi_{i=1}^k {r_i}^{e_i}$ (prime factorization of c)
Then, think of ${r_i}$ for all integers $i$ ($1 \le i \le k$)
Note that $\gcd(a,b)=1$.
So,note that if $r_i | a$ then $\gcd(b,r_i)=1$, and vice versa. ...(1)
$a \times b = c^n = \Pi_{i=1}^k {r_i}^{n \times e_i}$
Let $P= \{ p_i|1 \le i \le s, i \in \boldsymbol Z \}$,
$Q= \{ q_i|1 \le i \le t, i \in \boldsymbol Z \}$,
$R= \{ r_i|1 \le i \le k, i \in \boldsymbol Z \}$
then, $P \cup Q = R$ and $ P \cap Q = \varnothing$
Let $v_p(n)$ be the exponent of the largest power of p that divides n.
Then, $n|v_{r_i}(c^n)$ for all integers $i$ ($1 \le i \le k$) ...(2)
by (1),(2),
$n|v_{p_i}(a)$ for all integers $i$ ($1 \le i \le s$) and
$n|v_{q_i}(b)$ for all integers $i$ ($1 \le i \le t$)
So the proof is correct.
Best Answer
Note $\,(a\!-\!b)(a\!+\!c)=a^2\,$ and $\, \overbrace{(\color{#c00}{a\!-\!b,a\!+\!c})=1}^{\text{see comments}}\,$ hence $\,a\!-\!b\,$ is a square. $\ \small\rm QED$
Or $\,d \!=\! a\!-\!b\,$ below $\:\!\Rightarrow\:\! d = \color{#0a0}{(a,b)^2},\,$ i.e. use factorization refinement (Four Number Theorem).
Lemma $\,\ \color{c00}{cd = ab}\, \Rightarrow\, d = \color{#0a0}{(d,a)(d,b)}\ $ if $\ (a,b,c,d)\! =\! 1.\ $ Proof $ $ one liner: expand $\rm\color{#0a0}{product}$.