Analyzing the Expression I = a^2 + b^2 + c^2 with Consecutive Integers

This solution breaks down the problem of determining the properties of the expression I = a² + b² + c². We are given that a and b are consecutive integers, and c is their product (c = ab). We will use algebraic methods to find out if I is odd or even, and if it's a perfect square.

Understanding the Given Conditions

• The expression is defined as I = a² + b² + c².
• The key condition is that a and b are consecutive integers. This means they follow each other directly, like 3 and 4, or 7 and 8.
• We are also told that c = ab, meaning c is the product of these two consecutive integers.
• Our task is to characterize the value of I based on these rules.

Deriving the Properties of I Algebraically

To make the algebra easier, let's represent the consecutive integers using a base integer n. We can set a = n and b = n+1.

Now, let's find an expression for c using n:

\[ c = ab = n(n+1) \]

Substitute a, b, and c in terms of n into the expression for I:

\[ I = n^2 + (n+1)^2 + [n(n+1)]^2 \]

Let's expand the terms:

• (n+1)² = n² + 2n + 1
• [n(n+1)]² = (n² + n)² = (n²)² + 2(n²)(n) + n² = n⁴ + 2n³ + n²

Now, substitute these expanded forms back into the equation for I:

\[ I = n^2 + (n^2 + 2n + 1) + (n^4 + 2n^3 + n^2) \]

Combine the terms by degree:

\[ I = n^4 + 2n^3 + (n^2 + n^2 + n^2) + 2n + 1 \]

\[ I = n^4 + 2n^3 + 3n^2 + 2n + 1 \]

Identifying I as the Square of an Odd Integer

We have simplified I to n⁴ + 2n³ + 3n² + 2n + 1. Let's see if this is a perfect square.

Consider the expression (n² + n + 1). Let's square it:

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

Expanding this multiplication gives:

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

\[ = (n^4 + n^3 + n^2) + (n^3 + n^2 + n) + (n^2 + n + 1) \]

\[ = n^4 + (n^3 + n^3) + (n^2 + n^2 + n^2) + (n + n) + 1 \]

\[ = n^4 + 2n^3 + 3n^2 + 2n + 1 \]

This is exactly the expression we found for I! So, we can confidently say:

\[ I = (n^2 + n + 1)^2 \]

This shows that I is always a perfect square. Now, we need to determine if the number being squared, (n² + n + 1), is odd or even.

Let's look at the term n² + n:

• We can rewrite this as n(n+1).
• Since n and n+1 are consecutive integers, one of them must be even.
• The product of any integer and an even number is always even. Thus, n(n+1) is always even.
• Therefore, n² + n is an even number.

Now consider n² + n + 1:

\[ n^2 + n + 1 = (\text{even number}) + 1 \]

An even number plus 1 is always an odd number.

So, (n² + n + 1) is always an odd integer.

Since I = (n² + n + 1)², and (n² + n + 1) is always odd, I must be the square of an odd integer.

Comparing the Result with the Options

Let's evaluate the choices provided:

• Option 1: claims I is even and not a square. This is incorrect because I is the square of an odd number, making it odd, not even.
• Option 2: claims I is odd but not a square. This is incorrect because we proved I is always a perfect square.
• Option 3: claims I is the square of an even integer. This is incorrect because the base number, (n² + n + 1), is always odd.
• Option 4: claims I is the square of an odd integer. This matches our derivation perfectly.
• Option 5: This appears to be an incomplete or invalid option.

Final Conclusion

Through algebraic manipulation, we confirmed that I = a² + b² + c², where a and b are consecutive integers and c = ab, is equal to (n² + n + 1)². We also showed that (n² + n + 1) is always an odd integer. Therefore, I is the square of an odd integer.