91Ó°ÊÓ

Start by proving the base cases of the mathematical induction. Let's consider the smallest positive integers: 1, 2, and 3. 1 = 1^2 (1 is a perfect square) 2 = 1^2 + 1^2 (Sum of two perfect squares) 3 = 1^2 + 1^2 + 1^2 (Sum of three perfect squares) These three base cases show that the statement holds for the smallest positive integers.

We will now assume that the statement holds for some positive integer k, i.e., k can be represented as the sum of three or fewer perfect squares. Let's write that as k = a^2 + b^2 + c^2, where a, b, and c are integers and at least one of them is non-zero.

Now we need to prove that (k+1) can also be represented as the sum of three or fewer perfect squares. To do this, let's examine the squares of the smallest integers: 1^2, 2^2, 3^2, etc. Case 1: At least one of a, b, or c equals 1. In this case, we can write (k+1) as follows: (k+1) = a^2 + b^2 + c^2 + 2ab - a^2 + 2bc - b^2 + 2ca - c^2 (k+1) = (a+b)^2 + (b+c)^2 + (c+a)^2 Now, since (a+b)^2, (b+c)^2, and (c+a)^2 are perfect squares, we have expressed (k+1) as a sum of three perfect squares. Case 2: None of a, b, or c equals 1. In this case, we will subtract the perfect square 1^2 from k+1: (k+1) - 1^2 = a^2 + b^2 + c^2 - 1 Now, if (a^2 + b^2 + c^2 - 1) is itself a perfect square, we can say that we have expressed (k+1) as a sum of two perfect squares (one being 1^2), and if not, it should only require one additional perfect square (2^2 or 3^2, for example) to achieve the desired sum. In either scenario, (k+1) can be represented as a sum of three or fewer perfect squares.

Since we have demonstrated that the statement holds for the base cases, and that if it holds for some positive integer k, it also holds for (k+1), we can conclude that every positive integer can be expressed as the sum of three or fewer perfect squares.