Final answer:
To prove that 3a² - 1 is never a perfect square when a is an integer, we assume the opposite and use contradiction by considering the equation modulo 3. We show that the equation cannot hold, thus proving our assumption incorrect.
Step-by-step explanation:
To prove that 3a² - 1 is never a perfect square when a is an integer, we can use contradiction.
Let's assume that 3a² - 1 is a perfect square. Therefore, there exists an integer n such that n² = 3a² - 1.
We can rewrite the equation as n² - 3a² = -1.
Now, let's consider the equation modulo 3. Looking at the possible remainders when a number is divided by 3, we can see that the squares can only give remainders 0 or 1. But the left side of the equation gives a remainder of 2 when divided by 3, and the right side gives a remainder of -1 = 2, which is not possible.
Hence, our assumption was incorrect, and therefore 3a² - 1 is never a perfect square when a is an integer.