The question is about Possible pairs of Solutions. We need to find out the possible pairs which satisfies the given polynomial function. Polynomials is a simple topic which involves approx. two to three concepts. Make sure that you a get of hold of them by solving these questions. A range of CAT questions can be asked based on this simple concept. This question uses the concept of prime factorization from Number Theory.

Question 1: How many pairs of integer (a, b) are possible such that a^{2} – b^{2} = 288?

- 6
- 12
- 24
- 48

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288 = 2^{5} * 3^{2}. So it has 6 * 3 = 18 factors. Or, there are 9 ways of writing this number as a product of two positive integers.

Let us list these down.

1 * 288, 2 * 144, 3 * 96, 4 * 72, 6 * 48, 8 * 36, 9 * 32, 12 * 24 and 16 * 18

Now, this is where the question gets interesting. If a, b are integers either a + b and a – b have to be both odd or a + b and a – b have to be both even. So, within this set of possibilities

1 * 288, 3 * 96 and 9 * 32 will not result in integer values of a, b. So, there are 6 sets of numbers that work for us.

Moving on to these six sets; let us start with one example and see how many possibilities we can generate from this.

Let us consider the set 2 * 144.

Let us solve for this for a, b being natural numbers first, then we will extend this to integers.

When a, b are natural numbers

a + b > a – b

So, a + b = 144, a – b = 2; a = 73 and b = 71

Now, if a = 73, b = 71 holds good. We can see that a = 73, b = –71 also holds good. a = –73, b = 71 works and so does a = –73, b = –71.

There are 4 possibilities.

a = 73, b = 71

a = 73, b = –71

a = –73, b = 71

a = –73, b = –71

So, for each of the 6 products remaining, we will have 4 possibilities each. Total number of (a, b) that will satisfy this equation = 6 * 4 = 24.

The question is **"How many pairs of integer (a, b) are possible such that a ^{2} – b^{2} = 288?"**

Choice C is the correct answer.

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