1. ## Need to clarify a quadratic equation (simple)

Is it true that in this equation:

$\frac{x}{99}=(a-b)(a+b)$

To make the fraction a whole number (x being a multiple of 99) it is not possible to make the (a-b) and (a+b) make a multiple of 99, with a and b being single digit numbers only.

I don't think so myself. The only possible situation would be used two digit numbers e.g. a as 10 and b as 1. However this would defy the question.

2. Why would you want a- b or a+ b to be a multiple of 99? The left side is not likely to be!

The right side is $(a- b)(a+ b)= a^2- b^2$, a difference of squares.
Neither 1 nor 2 is a difference of squares but 3 is: you could have x= 3(99)= 297, a= 2, b= 1. 4 is not a difference of squares but 5 is: you could have x= 5(99)= 495, a= 3, b= 2.

3. Originally Posted by Mukilab
Is it true that in this equation:

$\frac{x}{99}=(a-b)(a+b)$

To make the fraction a whole number (x being a multiple of 99) it is not possible to make the (a-b) and (a+b) make a multiple of 99, with a and b being single digit numbers only.

I don't think so myself. The only possible situation would be used two digit numbers e.g. a as 10 and b as 1. However this would defy the question.
1. I assume that $x > 0, a > 0, b > 0$. Thus $a > b$.

2. If $a-b=1$ then $x = 99(a+b)$

3. Therefore you'll get these single digit results:

$(a,b) = (a, a-1),~ a \in \{2,3,...,9\}$

4. If $a-b=3$ then $x = 3 \cdot 99(a+b)$

5. Therefore you'll get these single digit results:

$(a,b) = (a, a-3),~ a \in \{4,5,...,9\}$

6. Use the same way to dtermine the values of a and b if

$a-b = 5$
$a-b = 7$

7. If $a-b=9$ then you have to use at least two-digit-numbers.