Find all the positive integers (x,y,z) such that xyz = 5(x + y + z)
If then .
We will show that if then . We will use induction starting at . At this statement is true by above. Say is true i.e. then . This completes induction.
This tells us that if then it must mean that and for there to be a possible solution. And if then there is a chance to be a solution. Since this is a symmetric equation it means if is a solution then any permutation is a solution. Therefore, it is sufficient to check whether this equation has integer solution for .
If then the equation is . We can write . Then add and subtract to get, , finally factor, . The possible solutions are and . But because we cannot choose the negative solutions. Thus, in fact the solutions are and (in that order). Because of permuting solutions we can ignore some and write: .
If then the equation is . Rewrite as . The solutions are (no negatives) and . Thus, and . This gives us .
If then equation is . Rewrite as . Thus, and . This gives and . This gives us .
And if then as explained above there are no other solutions if we think of permutations.