For how many unique coordinate points (P, Q), such that P and Q are in

This is a tough one! The first thing you should do is that we should factor the difference of squares P2 − Q2 into (P − Q)(P + Q). That gives us:

(P − Q)(P + Q) = 1155

That means that (P - Q) and (P + Q) must both be factors of 1155, since the product of (P − Q) and (P + Q) equals 1155. Because the product of (P − Q) and (P + Q) = 1155, if we can figure out the total number of factors 1155 has, we can figure out the number of possible values for P and Q, because every unique pair of values for (P − Q) and (P + Q) will correspond to a unique value pair for (P,Q).

To find the total number of factors in 1155, we need to take its prime factorization. 1155 ends in a 5, so we can take out a 5 to get:

5 * 231

The sum of the digits in 231 is 6, which is a number divisible by 3, so 3 must be a factor of 231. Breaking out a 3, we get:

5 * 3 * 77

Lastly, 77 = 7 * 11, so the prime factorization of 1155 is:

1155 = 3 * 5 * 7 * 11

Now we use the method from the Counting Factors of Large Numbers lesson to determine the total number of positive factors in 1155. Since there are no exponents written next to each prime factor, the exponent of each factor must be 1. We then add 1 to each number, giving 2, and then do:

2 * 2 * 2 * 2 = 16

So there are 16 total factors in 1155, which corresponds to 16 unique values for (P - Q) and (P + Q). Since (P − Q) and (P + Q) have 16 unique values, P and Q also must have 16 unique positive values. However, since (P − Q) and (P + Q) can both be negative as well, that accounts for another 16 unique values of P and Q. Because there are 16 positive and 16 unique value pairs for P and Q, the total number of unique coordinate points (P,Q) is 32.