$\displaystyle a.b-c=4$

$\displaystyle b.c-a=3$

in equality, a, b, c are positive integers.

What is total of $\displaystyle a+b+c $ minumum?

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- Aug 11th 2008, 05:11 AMOPETHa question......
$\displaystyle a.b-c=4$

$\displaystyle b.c-a=3$

in equality, a, b, c are positive integers.

What is total of $\displaystyle a+b+c $ minumum? - Aug 11th 2008, 05:23 AMQuick
$\displaystyle a.b-c=4$

$\displaystyle b.c-a=3$

add the two equations together:

$\displaystyle ab+bc-a-c=7$

move things around:

$\displaystyle ab-a+bc-c=7$

factor:

$\displaystyle a(b-1)+c(b-1)=7$

factor again:

$\displaystyle (a+c)(b-1)=7$

hmm, 7 is a prime number though and so cannot have two integer factors,__unless__$\displaystyle b-1 = 1$ and $\displaystyle a+c=7$__or__$\displaystyle b-1=7$ and $\displaystyle a+c=1$ (which means that in this case either a or c must be zero, I don't know if you consider that a positive integer)

In the first case: $\displaystyle b = 2$ and $\displaystyle a+c=7$

So if we add those two together: $\displaystyle a+b+c=9$

In the second case: $\displaystyle b = 8$ and $\displaystyle a+c=1$

So if we add those two together: $\displaystyle a+b+c=9$

So the answer is: $\displaystyle \boxed{a+b+c=9}$