Difference of inverse squares Given that the positive number $a$ is the difference of inverse squares:
$$a = \frac{1}{n^2} - \frac{1}{m^2}, m, n \in \mathbb{N},$$
could it well be that the $pa$ is also a difference of inverse squares , when p - some natural number ?

If 
$$a = \frac{1}{n^2} - \frac{1}{m^2},$$
where $m, n \in \mathbb{N}$ and $\mathbb{N}$ is the set of natural numbers, then
$$2a = \frac{2}{n^2} - \frac{2}{m^2}.$$
The OP is asking for solutions to the Diophantine equation
$$2a = \frac{2}{n^2} - \frac{2}{m^2} = \frac{1}{r^2} - \frac{1}{s^2},$$
where $r, s \in \mathbb{N}$.
This reduces to
$$2{r^2}{s^2}(m + n)(m - n) = {m^2}{n^2}(r + s)(r - s).$$
Consequently, we have:
$$2{r^2}{s^2} \mid {m^2}{n^2}(r + s)(r - s).$$
Assuming $\gcd(r, s) = 1$, then we have either:


(1) $2 \mid m$, or
(2) $2 \mid n$, or
(3) $2{r^2}{s^2} \mid {m^2}{n^2}$, or
(4) $2 \mid (r + s)(r - s)$.


Perhaps this could shed some light into Antony's answer.
Update - Antony gave the following solutions:
$$n  \hspace{0.1in} m  \hspace{0.1in}  r  \hspace{0.1in} s$$
$$5  \hspace{0.1in} 7  \hspace{0.1in}  5  \hspace{0.1in} 35$$
$$5  \hspace{0.1in} 10 \hspace{0.1in}  4  \hspace{0.1in} 20$$ 
$$6  \hspace{0.1in} 10 \hspace{0.1in}  5  \hspace{0.1in} 15$$
$$9  \hspace{0.1in} 11 \hspace{0.1in}  11 \hspace{0.1in} 99$$
$$10 \hspace{0.1in} 14 \hspace{0.1in}  10 \hspace{0.1in} 70$$
$$5  \hspace{0.1in} 15 \hspace{0.1in}  3  \hspace{0.1in} 5$$
Notice that all the known solutions (so far) satisfy:


(4) $2 \mid (r + s)(r - s)$.


