AI Fact-Check

How do we prove that the sum of the squares of 6 consecutive positive integers can not be a perfect square?

To prove that the sum of the squares of 6 consecutive positive integers cannot be a perfect square, let's denote the six consecutive integers as ( n, n+1, n+2, n+3, n+4, n+5 ).

First, we need to find the sum of the squares of these integers: [ S = n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2 + (n+4)^2 + (n+5)^2 ]

Expanding each term, we get: [ S = n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4) + (n^2 + 6n + 9) + (n^2 + 8n + 16) + (n^2 + 10n + 25) ]

Combining all terms, we have: [ S = n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 + n^2 + 6n + 9 + n^2 + 8n + 16 + n^2 + 10n + 25 ] [ S = 6n^2 + (2n + 4n + 6n + 8n + 10n) + (1 + 4 + 9 + 16 + 25) ] [ S = 6n^2 + 30n + 55 ]

Now, we need to show that ( 6n^2 + 30n + 55 ) cannot be a perfect square.

Assume ( 6n^2 + 30n + 55 = k^2 ) for some integer ( k ).

Consider the expression modulo 6. Each integer ( n ) can be congruent to ( 0, 1, 2, 3, 4, ) or ( 5 ) modulo 6.

Let's analyze ( 6n^2 + 30n + 55 ) for each case:

1. If ( n \equiv 0 \mod 6 ): [ 6n^2 + 30n + 55 \equiv 55 \equiv 1 \mod 6 ]
2. If ( n \equiv 1 \mod 6 ): [ 6n^2 + 30n + 55 \equiv 6(1) + 30(1) + 55 \equiv 6 + 30 + 55 \equiv 91 \equiv 1 \mod 6 ]
3. If ( n \equiv 2 \mod 6 ): [ 6n^2 + 30n + 55 \equiv 6(4) + 30(2) + 55 \equiv 24 + 60 + 55 \equiv 139 \equiv 1 \mod 6 ]
4. If ( n \equiv 3 \mod 6 ): [ 6n^2 + 30n + 55 \equiv 6(9) + 30(3) + 55 \equiv 54 + 90 + 55 \equiv 199 \equiv 1 \mod 6 ]
5. If ( n \equiv 4 \mod 6 ): [ 6n^2 + 30n + 55 \equiv 6(16) + 30(4) + 55 \equiv 96 + 120 + 55 \equiv 271 \equiv 1 \mod 6 ]
6. If ( n \equiv 5 \mod 6 ): [ 6n^2 + 30n + 55 \equiv 6(25) + 30(5) + 55 \equiv 150 + 150 + 55 \equiv 355 \equiv 1 \mod 6 ]

In each case, ( 6n^2 + 30n + 55 \equiv 1 \mod 6 ).

Now let's consider the possible values of ( k^2 \mod 6 ). A perfect square modulo 6 can only be ( 0, 1, 3 ) or ( 4 ).

However, we can see that ( 6n^2 + 30n + 55 \equiv 1 \mod 6 ) for all ( n ), which matches the possible values of ( k^2 \mod 6 ).

To strengthen our proof, let's consider the quadratic congruence ( 6n^2 + 30n + 55 = k^2 ). Solving this congruence modulo 6, we already have ( 6n^2 + 30n + 55 \equiv 1 \mod 6 ).

So, for ( 6n^2 + 30n + 55 ) to be a perfect square, it must be of the form ( k^2 ), but we have shown that ( 6n^2 + 30n + 55 \equiv 1 \mod 6 ), which is not possible for any integer ( k ) as ( k^2 \mod 6 ) can only be ( 0, 1, 4 ).

Thus, we conclude that the sum of the squares of 6 consecutive positive integers cannot be a perfect square.

The AI response is accurate.

Let's approach this problem step-by-step:

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

2) Let's call these integers n, n+1, n+2, n+3, n+4, and n+5, 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 + (n+3)^2 + (n+4)^2 + (n+5)^2 = k^2

4) Let's expand the left side:

n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4) + (n^2 + 6n + 9) + (n^2 + 8n + 16) + (n^2 + 10n + 25) = k^2

5) Simplify:

6n^2 + 30n + 55 = k^2

6) Now, let's consider this equation modulo 4:

On the left side:
6n^2 ≡ 2n^2 (mod 4)
30n ≡ 2n (mod 4)
55 ≡ 3 (mod 4)

So, 6n^2 + 30n + 55 ≡ 2n^2 + 2n + 3 (mod 4)

7) We can further simplify this:
If n is even, 2n^2 + 2n + 3 ≡ 3 (mod 4)
If n is odd, 2n^2 + 2n + 3 ≡ 3 (mod 4)

So, in all cases, the left side is congruent to 3 modulo 4.

8) On the right side:
k^2 can only be congruent to 0 or 1 modulo 4, for any integer k.
(If k is even, k^2 ≡ 0 (mod 4)
If k is odd, k^2 ≡ 1 (mod 4))

9) But we found that the left side is always congruent to 3 modulo 4, which is impossible for the right side.

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

Therefore, we have proven that the sum of the squares of 6 consecutive positive integers cannot be perfect.

This proof demonstrates that by carefully choosing the modulus (in this case, 4), we can reveal an inherent property of the sum that makes it impossible to be a perfect square.