To prove Cynthia's conjecture, let's restate it formally:

Conjecture: For any two consecutive natural numbers $n$ and $n+1$, the difference of the squares of these two numbers is equal to the sum of the two numbers.

Step 1: Define the Consecutive Integers

Let $n$ and $n+1$ be two consecutive natural numbers.

Step 2: Calculate the Difference of Their Squares

The difference of the squares of $n$ and $n+1$ is: $(n+1)^2 - n^2$

Expanding both squares: $(n+1)^2 = n^2 + 2n + 1$ $n^2 = n^2$ Thus: $(n+1)^2 - n^2 = (n^2 + 2n + 1) - n^2$

Simplifying, we get: $(n+1)^2 - n^2 = 2n + 1$

Step 3: Calculate the Sum of the Two Numbers

The sum of $n$ and $n+1$ is: $n + (n+1) = 2n + 1$

Step 4: Compare the Results

We see that: $(n+1)^2 - n^2 = 2n + 1$ and $n + (n+1) = 2n + 1$

Since both expressions are equal, we conclude that: $(n+1)^2 - n^2 = n + (n+1)$

Conclusion

Cynthia's conjecture is true. The difference of the squares of two consecutive natural numbers is indeed equal to the sum of those two numbers.

Would you like further details on this proof or have any additional questions?

Here are five questions to explore further:

1. How can this conjecture be used to check whether two numbers are consecutive?
2. What happens if we try this with non-consecutive integers?
3. Is there a similar pattern if we look at the difference of cubes of consecutive integers?
4. Does this property apply to non-natural numbers, like negative integers or zero?
5. How might this formula relate to properties of arithmetic sequences?

Tip: For consecutive numbers, many algebraic patterns emerge. Experimenting with differences and sums often reveals interesting relationships.