If two integers are chosen at random out of the set {2, 5, 7, 8}, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers?

Question

Correct Answer is 2/3, my answer is coming out as 1/3

Akshat · Accepted Answer

Since 4C2 = 6, we only have 6 ways to choose 2 numbers from the 4 numbers in the set. We can just list the pairs of number and their products:
2 x 5 = 10
2 x 7 = 14
2 x 8 = 16
5 x 7 = 35
5 x 8 = 40
7 x 8 = 56
Now we need to determine which of these products are in the form of a^2 - b^2 (difference of squares), where a and b are positive integers. Let's list the first 20 positive perfect squares:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, ...
We see that the difference of a pair of consecutive perfect squares is odd, and we see that 324 - 289 = 18^2 - 17^2 = 35, which is one of the products and the only odd product.
Now, since the remaining 5 products are even, they must be a difference of two non-adjacent perfect squares. Let's analyze each of these 5 products:
10: Since 9 - 1 = 8 and 16 - 4 = 12, we see that there is no way we can get a difference of 10.
14: Since 16 - 4 = 12 and 25 - 9 = 16, we see that there is no way we can get a difference of 14.
16: Since 25 - 9 = 16, we see that 16 = 5^2 - 3^2.
40: Since 121 - 81 = 16, we see that 40 = 11^2 - 9^2.
56: Since 225 - 169 = 56, we see that 56 = 15^2 - 13^2.
Since 4 of the 6 products can be written as a^2 - b^2, the probability is 4/6 = 2/3.