Given the following function: $$f(x) = \sqrt{ (2155 - 6x)^2-4x}$$ where x is an integer and the function also generates an integer value, is there an algorithm to determine its integer solutions?
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First, note that for most positive $x$, $(2155-6x)^2-4x>(2155-6x-1)^2$. Similarly, for most negative $x$, we can place a bound. So you only need to check a finite number of $x$ at worst. – Thomas Andrews Feb 07 '15 at 13:33
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HINT : If you want to find $(x,y)\in\mathbb Z$ such that $$y^2=(2155-6x)^2-4x,$$ since we have $$36 x^2-25864 x+4644025-y^2=0\Rightarrow x=\frac{6466\pm\sqrt{9y^2+12931}}{18},$$ we know that $9y^2+12931$ has to be a perfect square.
Hence, for some $z\in\mathbb Z$, we have $$9y^2+12931=z^2\iff (z-3y)(z+3y)=67\times 193$$ where $67,193$ are primes. Then, you can find $(y,z)$ to get $(x,y)$.
mathlove
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@Nominal: Interestingly, your questions can be solved by finding $(X,Y)$ from $f(X,Y)\times g(X,Y)=N$ for some $N\in\mathbb Z$. So, you can find $(y,z)$ from $(z-3y)(z+3y)=67\times 193$, can't you? – mathlove Feb 08 '15 at 10:55