One Equation, Five Variables?

Relevant wiki: Quadratic Diophantine Equations - Solve by Bounding Values

Without loss of generality, assume $a\geqslant b\geqslant c\geqslant d\geqslant e.$

$\therefore 5a\geqslant a+b+c+d+e=abcde.$

However, when $5a=a+b+c+d+e=abcde$ , it means $a=b=c=d=e$ , which is invalid, since if it is true, then $5a=a^{5}$ . This implies $5=a^4$ , then $a$ is not a positive integer, contradiction occurs.

$\therefore 5a\neq abcde,$

$\therefore 5a> abcde\Rightarrow bcde< 5.$

$\because a,b,c,d,e\textrm{ are all positive integers},$

$\therefore bcde\leqslant 4.$

$\therefore \textrm{The possible parings of }b,c,d,e\textrm{ are}:$

$(1,1,1,1),\; (2,1,1,1),\; (2,2,1,1)\; (3,1,1,1).$

Substitute to the original equation, giving

$[1].\textrm{ For the paring }(1,1,1,1),\textrm{ we have }a+4=a,\textrm{ which is invalid}.$

$[2].\textrm{ For the paring }(2,1,1,1),\textrm{ we have }a+5=2a\Rightarrow a=5.$

$[3].\textrm{ For the paring }(2,2,1,1),\textrm{ we have }a+6=4a\Rightarrow a=2.$

$[4].\textrm{ For the paring }(3,1,1,1),\textrm{ we have }a+6=3a\Rightarrow a=3.$

$\therefore \textrm{There are only } \boxed{3} \textrm{ pairs which are valid, which are }$

$\left\{\begin{matrix} a_{1}=5\\ b_{1}=2\\ c_{1}=1\\ d_{1}=1\\ e_{1}=1 \end{matrix}\right.$ , $\; \; \left\{\begin{matrix} a_{2}=2\\ b_{2}=2\\ c_{2}=2\\ d_{2}=1\\ e_{2}=1 \end{matrix}\right.$ , $\; \; \left\{\begin{matrix} a_{3}=3\\ b_{3}=3\\ c_{3}=1\\ d_{3}=1\\ e_{3}=1 \end{matrix}\right.$ .

Can anyone link any website so I can look for the theory related to this, though I found the answer but need to look again. Thanx