How do I prove the following claim

If a perfect square is divided by 4, then the remainder is either 0 or 1.

(2k + 1)^2 = 4k^2 + 4k + 1 gives a remainder of 1
(2k)^2 = 4k^2 gives a remainder of 0

Well, I'd do a proof by counterexample. We show that a remainder of 2 or 3 is impossible.

Ok, if it's the product of two squares, the last digit can be:
0, 1, 4, 5, 6, or 9. Nothing else. Let's consider all squares up to 100. All of them check.

Something is divisible by four if the last two digits is divisible by 4.

All of our number now are going to be two digits or higher. So, let's get formal: a number n is made
up of x+d, where x is everything but the last digit d, number ranging from 0 to 9.

So the square is going to be something like x^2 +2xd + d^2

For the cases, you'll get:

x^2 (d = 0)
x^2 +2x + 1 (d = 1)
x^2 +4x + 4 (d = 2)
x^2 +6x + 9 (d = 3)
x^2 +8x + 16 (d = 4)
x^2 +10x + 25
x^2 + 12x + 36
x^2 + 14x + 49
x^2 + 16x + 64
x^2 + 18x + 81 (d= 9)

Now, recall that x is always going to be a multiple of 10, x^2 is a multiple of 100, so these will always be divisible by 4. The linear terms always have an even coefficient, and x is a multiple of 10, so the middle terms are always divisible by four. So you look to the last terms and see that the remainder is always zero or one.

This was an interesting exercise. Thanks.