prove that the sum of two consecutive perfect squares is always an odd number

To prove that the sum of two consecutive perfect squares is always an odd number, let's work through the problem step by step:

### Step 1: Represent consecutive perfect squares
Let the two consecutive integers be \( n \) and \( n+1 \).

Their perfect squares would be:
\[
n^2 \quad \text{and} \quad (n+1)^2
\]

### Step 2: Sum the two consecutive perfect squares
Now, compute the sum of these two perfect squares:
\[
n^2 + (n+1)^2
\]
Expanding the second term:
\[
n^2 + (n+1)^2 = n^2 + (n^2 + 2n + 1)
\]
Simplifying:
\[
n^2 + n^2 + 2n + 1 = 2n^2 + 2n + 1
\]

### Step 3: Analyze the result
The expression \( 2n^2 + 2n + 1 \) is clearly an odd number because it has the form:
\[
2(\text{some integer}) + 1
\]
Any number of the form \( 2k + 1 \), where \( k \) is an integer, is an odd number.

### Conclusion
Thus, the sum of two consecutive perfect squares is always an odd number.

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Would you like more details or have any questions about this proof?

Here are some related questions to explore:
1. What happens if the numbers are not consecutive perfect squares?
2. Is the sum of three consecutive perfect squares also odd?
3. How would the result change if negative integers were involved?
4. Can we generalize this for consecutive perfect cubes?
5. Is the product of two consecutive perfect squares odd or even?

**Tip:** When simplifying expressions, always check if the result fits a general form like \( 2k+1 \) for odd numbers.

Prove that the sum of two consecutive perfect squares is always an odd number.

This detailed proof explains why the sum of two consecutive perfect squares is always an odd number, using algebraic expressions and the properties of integers. Perfect for students in Grades 8-10, the solution walks through each step clearly, highlighting key formulas and concepts in number theory.

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