Ask Your Question

Revision history [back]

click to hide/show revision 1
initial version

The given equation can be rewritten as $$3(2y+1)(2z+1)=(2x+1)(2w+1),$$ which suggests that solutions can be generated by first fixing values of $y,z$ and thus $n:=3(2y+1)(2z+1)$, then running $d$ over the divisors of $n$, and setting $x=\frac{d-1}2$ and $w=\frac{n/d-1}2$ .