MHB Math History question - Fibonacci Proof

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The discussion revolves around proving that the sum of two consecutive integers results in a square, specifically that if n + (n + 1) = h^2, then h is odd. Participants are exploring the relationship between the larger integer's square and the sum of nonzero squares. The initial steps involve recognizing that the sum can be expressed as 2n + 1, which is a perfect square. The conversation highlights the need for a structured approach to the proof, emphasizing the mathematical properties of odd integers and squares. The thread seeks clarity on how to proceed with the proof effectively.
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Can someone please help me with the following:

Prove that if the sum of two consecutive intergers is a square than the square of the larger integer will equal the sum of the nonzero squares.

Hint: if n+(n-1) = h^2 then h is odd.

Not really sure where to start.

Thanks in advance
 
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Well, $n + (n + 1) = 2n + 1$ is a square by what is given and $(n + 1)^2$ is ... ?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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