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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 ?

Antony
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Andrew
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3 Answers3

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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)

Antony
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  • @Galc127 , for numbers less that 100 my program had found more that 20 solutions . – Antony Oct 14 '14 at 11:20
  • do you mind to share your thoughts? – Galc127 Oct 14 '14 at 11:21
  • 5 7 5 35 ;
    5 10 4 20 ;
    6 10 5 15 ;
    9 11 11 99 ;
    10 14 10 70 ;
    5 15 3 5. This is first 6 answers in format (m,n,k,p) where $2(1/(mm)-1/(nn))=1/(kk)-1/(pp)$ . But I have no idea how to get formula     – Antony Oct 14 '14 at 11:23
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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

<p>(2) $2 \mid n$, or</p>

<p>(3) $2{r^2}{s^2} \mid {m^2}{n^2}$, or</p>

<p>(4) $2 \mid (r + s)(r - s)$.</p>

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

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

Bhaskar Vashishth
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