MHB Math History question - Fibonacci Proof

Thread starter wreader
Start date Feb 7, 2014
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History 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.

Feb 7, 2014

wreader

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

Last edited: Feb 7, 2014

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Feb 7, 2014

mathbalarka

Well, $n + (n + 1) = 2n + 1$ is a square by what is given and $(n + 1)^2$ is ... ?

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