How to solve the Diophantine equation xyz = xy + yz + zx?

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The video illustrates step-by-step how to solve the Diophantine equation xyz = xy + yz + zx.
As x, y and z are positive integers, so x, y, z≠0. By dividing both sides of the original equation, we have 1=1/x+1/y+1/z. The next step is to narrow down the range of possible integers, for example, by assuming x ≥ y ≥ z.

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Nice problems and well explained.
For the additional problem, (a, b, c, d)=(42, 7, 3, 2), (24, 8, 3, 2), (18, 9, 3, 2), (15, 10, 3, 2), (12, 12, 3, 2) and many others.

markkerbyson

I miss to write conditions for one of the problems in the last comment. Regarding "Q4-(10)",
a+b+c+d+e = a^3+b^3+c^3+d^3+e^3 = 5 (a, b, c, d, e: integers) (a≧b≧c≧d≧e)
Find all (a, b, c, d, e). Find more than 6 solutions. (If there are less than 6 solutions, prove it.)
is the correct problem. The condition (a≧b≧c≧d≧e) was omitted. [ Excuse me. ]

sy

For the additional problem, there are too many solutions. I list a few example solutions (a, b, c, d)=(8, 8, 4, 2), (12, 6, 4, 2), (20, 5, 4, 2)

kkyoung

How to solve the Diophantine equation xyz = xy + yz + zx?

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