Difference of inverse squares

Question

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 ?

Answer 1

Yes
$1/(5*5)-1/(7*7)=24/(25*49)$
$1/(5*5)-1/(35*35)=48/(25*49)$ so we have $2*((1/5)^2-(1/7)^2)=(1/5)^2-(1/35)^2$
But it very interesting to see formula (I've made program to get this answer)

Answer 2

$2a=1/(n/\sqrt{2})^2-1/(m/\sqrt{2})^2$, but $(n/\sqrt{2}) ,(m/\sqrt{2})$ won't be natural

Answer 3

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