# Is there any possible way to solve this equation with two different variables with different degrees? [closed]

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

## closed as unclear what you're asking by Dietrich Burde, Shailesh, Zain Patel, Leucippus, Claude LeiboviciAug 8 '17 at 15:19

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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 8 '17 at 14:52

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.