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$$649 + 96y = k^{2}$$ Also, $(y)$ or $(y + 1)$ or $(y-1)$ must be a perfect square. $y$ and $k$ are both natural numbers.

Aryaman Velampalli
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    Is this to be solved over the integers? Saying a perfect square seems to imply it. Write $y=m^2$ and similarly for the other two cases. – Ross Millikan Aug 08 '17 at 14:52

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I asked Alpha to solve $649+96x^2=y^2$ over the integers, then replaced $649$ with $553$ and $745$. Only $745$ gave solutions. The lowest is $x=8,y=83$ and there are a few families. It clearly used the continued fraction approach to find them. The usual approach to show there are no solutions is to find a clever modulus to show the equation is impossible.

Ross Millikan
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  • Hey there, I actually had a doubt in solving a question on Quora. If you can, will you please check my answer here: https://www.quora.com/Three-numbers-in-AP-are-removed-from-first-n-natural-numbers-Mean-of-the-remaining-numbers-is-43-4-What-is-n-if-one-of-the-removed-numbers-is-a-perfect-square – Aryaman Velampalli Aug 08 '17 at 15:14