To prove that the sum of the squares of three consecutive positive numbers cannot be a perfect square, we can use algebraic manipulation and properties of integers. Let's denote the three consecutive positive numbers as ( n-1 ), ( n ), and ( n+1 ), where ( n ) is a positive integer.

First, write the sum of the squares of these numbers:

[ (n-1)^2 + n^2 + (n+1)^2 ]

Next, expand and simplify each term:

[ (n-1)^2 = n^2 - 2n + 1 ] [ n^2 = n^2 ] [ (n+1)^2 = n^2 + 2n + 1 ]

Now, add these expressions together:

[ (n^2 - 2n + 1) + n^2 + (n^2 + 2n + 1) ]

Combine like terms:

[ n^2 - 2n + 1 + n^2 + n^2 + 2n + 1 = 3n^2 + 2 ]

So, the sum of the squares of three consecutive positive numbers is:

[ 3n^2 + 2 ]

To show that this expression cannot be a perfect square, assume for contradiction that it is a perfect square. Let:

[ 3n^2 + 2 = k^2 ]

for some integer ( k ). Rearrange this equation:

[ k^2 - 3n^2 = 2 ]

This is a type of Diophantine equation known as a Pell-like equation. We need to examine if there are any integer solutions to this equation.

Consider the equation modulo 3:

[ k^2 \equiv 2 \pmod{3} ]

Notice that the quadratic residues modulo 3 (i.e., the possible values of ( k^2 \mod 3 )) are 0 and 1. This means ( k^2 ) modulo 3 cannot be 2:

[ 0^2 \equiv 0 \pmod{3} ] [ 1^2 \equiv 1 \pmod{3} ] [ 2^2 \equiv 1 \pmod{3} ]

Thus, ( k^2 \equiv 2 \pmod{3} ) has no solutions, and therefore, it is impossible for ( 3n^2 + 2 ) to be a perfect square.

Hence, the sum of the squares of three consecutive positive numbers ( 3n^2 + 2 ) can never be a perfect square. This completes the proof.

The AI-generated answer is correct. Below, you will find a list of steps to clarify the process:

**Step-by-step approach using proof by contradiction**:

1) Let's assume that three consecutive positive integers exist whose sum of squares is a perfect square.

2) Let's call these integers n, n+1, and n+2, where n is a positive integer.

3) If our assumption is correct, then there must exist some integer k such that:

n^2 + (n+1)^2 + (n+2)^2 = k^2

4) Let's expand the left side:

n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4) = k^2

5) Simplify:

3n^2 + 6n + 5 = k^2

6) Now, let's consider this equation modulo 3 (i.e., consider the remainder when both sides are divided by 3):

On the left side:

3n^2 ≡ 0 (mod 3)

6n ≡ 0 (mod 3)

5 ≡ 2 (mod 3)

So, 3n^2 + 6n + 5 ≡ 2 (mod 3)

7) On the right side:

k^2 can only be congruent to 0 or 1 modulo 3, for any integer k.

(If k ≡ 0 (mod 3), then k^2 ≡ 0 (mod 3)

If k ≡ 1 (mod 3), then k^2 ≡ 1 (mod 3)

If k ≡ 2 (mod 3), then k^2 ≡ 1 (mod 3))

8) But we found that the left side is congruent to 2 modulo 3, which is impossible for the right side.

9) This contradiction proves that our initial assumption must be false.

Therefore, we have proven that the sum of the squares of 3 consecutive positive integers cannot be perfect.

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