I am trying to find all strictely positive integers $x,y$ so that $$z=\dfrac{4x+1}{4y-1}$$ is a strictely positive integer.

Some cases are possible when choosing special values of $x$ and $y$. However, I am not able to find the general solution.


(This is not the full answer), but we have at least infinite pairs $(x,y)\in \mathbb{N}^2$ for which $z$ is positive integer. Say $y$ is arbitrary, then $x=4ty-t-y$ will solve the equation where $t$ is arbitrary (positive) integer.

Actually, I just realized that are all solutions of the given equation.

Rewrite like this $$4zy-z=4x+1\Longrightarrow 4\mid z+1 \Longrightarrow z =4t-1$$ So we get $x=4ty-y-t\;\;\;\;(*)$. So, however I chose $t$ and $y$ and $x$ is of form $(*)$, then $z$ will be (positive) integer.


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