# Diophantus equation

• August 13th 2012, 05:51 AM
albert940410
Diophantus equation
please solve the following "Diophantus equation "

(I know how to solve it , may be your solution is better ,please have a try ,many thanks !)

mn+nr+mr=2(m+n+r)

where m,n,r are all positive integers

Ans:there are 7 combinations of (m,n,r)
• August 13th 2012, 08:34 AM
richard1234
Re: Diophantus equation
Move everything to one side and regroup:

$mn + nr + mr - 2m - 2n - 2r = 0$

$(mn - m - n) + (nr - n - r) + (mr - m - r) = 0$

This can be rewritten as

$(m-1)(n-1) + (n-1)(r-1) + (m-1)(r-1) = 3$

I'll let a = m-1, b = n-1, c = r-1 where a,b,c are non-negative integers, that is $ab + bc + ca = 3$.

Here, either (a,b,c) = (1,1,1) or (0,1,3) (up to re-arranging). This leaves 1! + 3! = 7 solutions.
• August 13th 2012, 06:21 PM
albert940410
Re: Diophantus equation
Richard1234:
Thank you very much !
a very good solution indeed !