Question

Find the general expression for three positive integers in arithmetic progression, and such that the sum of every two is a perfect square.

Answer 1

Denote the integers by $x-y,x,x+y$; and let
$2x-y=p^2,2x=q^2,2x+y=r^2$;
then $p^2+r^2=2q^2$,
or $r^2-q^2=q^2-p^2$.
This equation is satisfied by the suppositions,
$m(r-q)=n(q-p)$, $n(r+q)=m(q+p)$,
where $m$ and $n$ are positive integers.
From these equations we obtain by cross multiplication
$\frac{p}{n^2+2mn-m^2}=\frac{q}{m^2+n^2}=\frac{r}{m^2+2mn-n^2}$.
Hence we may take for the general solution
$p=n^2+2mn-m^2$, $q=m^2+n^2$, $r=m^2+2mn-n^2$;
whence $x=\frac{1}{2}{(m^2+n^2)}^2$, $y=4mn(m^2-n^2)$,
and the three integers can be found.
From the value of $x$ it is clear that $m$ and $n$ are either both even or both odd; also their values must be such that $x$ is greater than $y$, that is,
${(m^2+n^2)}^2>8mn(m^2-n^2)$,
or $m^3(m-8n)+2m^2{n}^2+8m{n}^3+n^4>0$;
which condition is satisfied if $m>8n$.
If $m=9$, $n=1$, then $x=3362$, $y=2880$, and the numbers are $482$, $3362$, $6242$. The sums of these taken in pairs are $3844$, $6724$, $9604$, which are the squares of $62$, $82$, $98$ respectively.