Bilal chooses two distinct positive integers. He adds the product of the integers to the sum of the integers, and then adds \(1\). He finds that the result is equal to \(196\).
Determine all possible pairs of integers that Bilal could have chosen.
Let \(x\) and \(y\) represent the two positive integers
that Bilal chooses. Since the integers are distinct, \(x\not =y\). Let \(x<y\). That is, let \(x\) represent the smaller of the two
integers.
The product of the two integers is \(xy\) and the sum is \((x+y)\).
Bilal adds the product of the numbers to the sum of the numbers and then
adds \(1\), and the result is \(196\). Thus, \[xy + x+ y + 1 = 196\] Factoring the left
side, by grouping the first two terms and the last two terms, we get
\[\begin{aligned}
x(y+1)+1(y+1)&=196\\
(x+1)(y+1)&=196
\end{aligned}\]
Since \(x\) and \(y\) are positive integers, then \(x+1\) and \(y+1\) are positive integers. Thus, we are looking for a pair of positive integers whose product is \(196\). There are four ways to factor \(196\) as a product of two positive integers: \[196=1\times 196=2\times 98=4\times 49=7\times 28=14\times 14\]
For the product \(196=1\times 196\), we have \(x+1 = 1\) and \(y+1 = 196\). Thus, \(x=0\) and \(y=195\). Since the required numbers are positive integers, this solution is inadmissible.
For the product \(196=2\times 98\), we have \(x+1 = 2\) and \(y+1 = 98\). Thus, \(x=1\) and \(y=97\). This is a valid solution.
For the product \(196=4\times 49\), we have \(x+1 = 4\) and \(y+1 = 49\). Thus, \(x=3\) and \(y=48\). This is a valid solution.
For the product \(196=7\times 28\), we have \(x+1 = 7\) and \(y+1 = 28\). Thus, \(x=6\) and \(y=27\). This is a valid solution.
For the product \(196=14\times 14\), we have \(x+1 = 14\) and \(y+1 = 14\). Thus, \(x=13\) and \(y=13\). Since the required numbers are distinct, this solution is inadmissible.
Therefore, there are three pairs of distinct positive integers that Bilal could have chosen: \(1\) and \(97\), \(3\) and \(48\), or \(6\) and \(27\). It can be shown that these three pairs do indeed each give the required result.