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.