How many pairs of integer (a, b) are possible such that a ² – b ² = 288

Explanatory Answer :

we know ,288 = 2^5 * 3^2.

Thus, there are (5+1) * (2+1) = 18 factors of 288 listed as following products:

1) 1 * 288,

2) 2 * 144,

3) 3 * 96,

4) 4 * 72,

5) 6 * 48,

6) 8 * 36,

7) 9 * 32,

8) 12 * 24, and

9) 16 * 18

a^2 – b^2 = (a-b).(a+b)

So, if both a and b are integers, (a+b) and (a–b) must be both odd or both even. from the above mentioned lists, 1 * 288, 3 * 96, and 9 * 32 don't satisfy such condition, but the remaining 6 product sets do:

1) 2 * 144,

2) 4 * 72,

3) 6 * 48,

4) 8 * 36,

5) 12 * 24, and

6) 16 * 18

Now, take an example of 16*18=(a-b).(a+b). There are 4 possible combination of (a, b) for 16*18 alone:

I. a=17, b=1

II. a=-17, b=-1

III. a=-17, b=1

IV. a=17, b=-1

So, for each of the remaining 6 product sets, we will have 4 possible combination. THUS, total number of (a, b) that will satisfy this equation = 6 * 4 = 24.

so ,the correct option is C