Proving $ \left\{\frac{y^2+z^2+2}{yz}\ \mid\ y,z\in\mathbb{N}\right\}\cap\mathbb{N}=\{4\} $

Question

I conjecture that: $$ \left\{\frac{y^2+z^2+2}{yz}\ \mid\ y,z\in\mathbb{N}\right\}\cap\mathbb{N}=\{4\} $$ Is this true? If yes, how to prove it?

Answer 1

Solution with the Vieta Jumping Technique

Let $k$ be an integer such that $$k=\dfrac{y^2+z^2+2}{yz}\tag{*}$$ for some positive integers $y$ and $z$. Let $(y,z)=(u,v)$ be a positive-integer solution to (*) such that $u\leq v$ and $u+v$ is smallest possible. Then, we have that $$(y,z)=\left(ku-v,u\right)=\left(\frac{u^2+2}{v},u\right)$$ is also a positive-integer solution to (*). However, due to the minimality of $(u,v)$, we must have $$\frac{u^2+2}{v}\geq v\,.$$ Therefore, $u\leq v\leq u+2$.

If $v=u$, then it is obvious that $(u,v)=(1,1)$, and so $k=4$. If $v=u+1$, then we have that the even number $uv=u(u+1)$ must divide the odd number $u^2+v^2+1=2u^2+2u+3$, which is a contradiction. If $v=u+2$, then $(u,v)=(1,3)$ is the only possibility, but this also gives $k=4$ and a contradiction as this pair is not minimal for this $k$.

Let $\left(a_j\right)_{j=0}^\infty$ be the sequence of integers given by $a_0=1$, $a_1=1$, and for $j=2,3,\ldots$, $$a_j=4\,a_{j-1}-a_{j-2}\,.$$ For examples, $a_2=3$, $a_3=11$, and $a_4=41$. More generally, $$a_j=\frac{(1+\sqrt{3})^{2j-1}-(1-\sqrt{3})^{2j-1}}{2^j\sqrt{3}}$$ for every $j=0,1,2,\ldots$. It can be shown that all solutions to (*) with $k=4$ are given by $$(y,z)=\left(a_j,a_{j+1}\right)\text{ and }(y,z)=\left(a_{j+1},a_j\right)$$ for $j=0,1,2,\ldots$.

Answer 2

This is precisely the type of question that Viète jumping solves. $$\frac{y^2+z^2+2}{yz}=k$$ If $y=z$, $\frac{2y^2+2}{y^2}=2+\frac2{y^2}$ and $y$ can only be 1, in which case $k=4$. Now let $y>z$, fix $z$ and $k$ and let $y$ vary as a variable $q$; we may write $$q^2-kz(q)+(z^2+2)=0$$ One root of this is $y$ and the other is $$q'=kz-y=\frac{z^2+2}y$$ so $q'$ is a positive integer. That $y>z$ implies that $q'=\frac{z^2+2}y<z$ as long as $z>2$. Hence $q'$ may replace $y$ to obtain a smaller valid solution.

Because of Viète jumping, now only the $z=1$ and $z=2$ cases need be considered.

• If $z=1$, $\frac{y^2+3}{y}=y+\frac3y=k$ and $y$ can only be 3, whence $k=4$.
• If $z=2$, $\frac{y^2+6}{2y}=\frac y2+\frac3y=k$ and a contradiction is reached because the first term implies an even $y$ and the second an odd $y$.

Hence $k$ is always 4 when it is a natural number.