Answer:
Let's suppose that the integer is x.
We know that x leaves a remainder of 1 when divided by 4, so we can express x as:
x = 4n + 1
where n is some integer.
Now we need to find the remainder when x^2 is divided by 8:
x^2 = (4n + 1)^2
x^2 = 16n^2 + 8n + 1
Notice that the first two terms (16n^2 + 8n) are divisible by 8, so we can ignore them.
The remainder when x^2 is divided by 8 is the same as the remainder of 1 when divided by 8:
x^2 ≡ 1 (mod 8)
Therefore, the remainder when the square of the integer is divided by 8 is 1.