2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

(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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.