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 ... ?
 

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