Math History question - Fibonacci Proof

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The discussion centers on proving that the sum of two consecutive integers results in a square, specifically that if \( n + (n + 1) = h^2 \), then the square of the larger integer \( (n + 1)^2 \) equals the sum of the nonzero squares of the integers involved. The key insight is that \( 2n + 1 \) is a perfect square, indicating that \( h \) must be odd. This establishes a foundational relationship between consecutive integers and their squares.

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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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