Question-1

**johnerdem** · April 24th 2008, 07:48 AM

Question says;

If A and B are positive numbers

How many different numbers can a be?

**Moo** · April 24th 2008, 07:51 AM

Hello,

$a^2-b^2=(a-b)(a+b)$

How many possibilities are there for a and b, being positive ?

**CaptainBlack** · April 24th 2008, 01:54 PM

Originally Posted by johnerdem

Question says;

If A and B are positive numbers

How many different numbers can a be?

You have $\frac{a^2-b^2}{a-b}=10,\ a,b \in \mathbb{Z}_+$

So (as $a \ne b$ ):

$a^2-b^2=10(a-b)\ \ \ \ \ ...(1)$

but $a^2-b^2=(a+b)(a-b)$ , so $(1)$ becomes:

$a+b=10\ \ \ \ \ ...(1)$

Now just enumerate the posibilities for $a$

RonL

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