Solve the following question and mark the best possible option.

Consider positive integers x and y such that the difference between x^{2 }+ y and x + y^{2} is a prime number. How many different values can the ordered pair (x, y) take?

#### Options

1

2

5

infinitely many

#### Solution

Suppose x^{2} + y > x + y^{2}. Let the difference between them be a prime number z so that z = (x^{2} + y) – (x + y^{2}) = (x – y) (x + y – 1). Since (x – y) < (x + y – 1), we get (x – y) = 1 and (x + y – 1) = z. Adding these two expressions, we get z + 1 = 2x – 1 ⇒x = `(z/2)` + 1. Since we know that x is an integer, we can conclude that (z/2) is an integer. In other words, z = 2, the only even prime number. This works when (x, y) could be (1, 2) or (2, 1). Thus, (x, y) can take 2 values.

**Alternate solution:** If both x and y are odd, then both expressions are even and their difference will be even. If both x and y are even, then both expressions are even and their difference is even. If one of x and y is odd and the other is even, then both expressions are odd and their difference is even. In all cases, the difference between the two expressions is even. Since this difference is a prime number, the only possibility is 2 (the only even prime number). With a little bit of trial and error, we can easily determine, that (x, y) could be (1, 2) or (2, 1). Thus, there are **2 values**.