Question

It is known that the equation x^2 + (x + k)^2= y^2 has positive integers solutions (x, y), where x and y are relatively prime. If k is a positive integer greater than 1, what is the minimum value of k?

a) 2
b) 3
c) 4
d) 5

Answer 1

The minimum value of k in the equation x^2 + (x + k)^2 = y^2 is 2.

Step-by-step explanation:

The minimum value of k in the equation x^2 + (x + k)^2 = y^2 is 2.

To find this, we can start by expanding the equation:

x^2 + x^2 + 2kx + k^2 = y^2

Combining like terms, we get:

2x^2 + 2kx + k^2 = y^2

We notice that the left side of the equation is always even, while the right side can be either even or odd. Therefore, for the equation to have positive integer solutions (x, y), k must be even. The smallest even number greater than 1 is 2, so the minimum value of k is 2.