Polynomials

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Tough one to crack!!

Polynomials - Possible pairs of Solutions

    How many pairs of integer (a, b) are possible such that a2 – b2 = 288?
    1. 6
    2. 12
    3. 24
    4. 48

 

  • Correct Answer
    Choice C. 24

Explanatory Answer

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Detailed Solution

288 = 25 × 32. 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.

Answer choice (c)

Correct Answer: 24.



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