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)

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- Aug 13th 2012, 05:51 AMalbert940410Diophantus 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) - Aug 13th 2012, 08:34 AMrichard1234Re: Diophantus equation
Move everything to one side and regroup:

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

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

This can be rewritten as

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

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

Here, either (a,b,c) = (1,1,1) or (0,1,3) (up to re-arranging). This leaves 1! + 3! = 7 solutions. - Aug 13th 2012, 06:21 PMalbert940410Re: Diophantus equation
Richard1234:

Thank you very much !

a very good solution indeed !