# Is there some exact form of $n$ for the number of $(i,j,k)$ satisfying $ijk = (n-i)(n-j)(n-k)$

For any positive integer $n$, $i,j,k$ are also positive integers, and $0 <i,j,k < n$. How many solutions of the form $(i,j,k)$ are there for the equation $ijk = (n-i)(n-j)(n-k)$?

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Let's get some trivial solutions out of the way. $n=2m$, $i=m-r$, $j=m$, $k=m+r$. Do you have a large supply of other solutions? –  Gerry Myerson Jun 2 '12 at 12:43
Here's another class: $y-1=(x-1)(x-2)/2$ then $n=xy,i=y,j=2y,k=n-x$, e.g. $(n,i,j,k)=(35,7,14,30)$. More generally if $y-1=(x-a)(x-b)/ab$ then $(n,i,j,k)=(xy,ay,by,xy-x)$ is a solution. e.g. $a=2,b=3,x=18,y=41$ and $(n,i,j,k)=(738,82,123,720)$. Also if $(n,i,j,k)$ then $(cn,ci,cj,ck)$ for positive integer $c$. –  Zander Jun 2 '12 at 13:57
If $i=j=k$ then $(2i,i,i,i)$ is a solution for all $i$. Set $n=i+j$ we get the another solution set $(i+j, i, j,\frac{i+j}{2})$. Replacing even $i$ & $j$ we get one set $(2i+2j, 2i, 2j, i+j)$. For both odd we get another $(2i+2j+2, 2i+1,2j+1, i+j+1)$ –  dear chap Jul 11 at 22:29