How to show that $\frac{a + b}{\gcd(a,b)^2}$ is a Fibonacci number when $ \frac{a+1}{b}+\frac{b+1}{a}$ is an integer? 
Let $a, b$ be positive integers such that the number $ \dfrac{a + 1}{b} + \dfrac{b + 1}{a}$ is also integer. Then, show that $\dfrac{a + b}{\gcd{(a, b)^{2}}}$ is a Fibonacci number.

I prove that : $ \displaystyle{\forall a, b \in \mathbb{N}} $ such that $ \displaystyle{ k = \frac{a + 1}{b} + \frac{b + 1}{a} \in \mathbb{N} \Rightarrow k = 3 }$ or $ \displaystyle{ k = 4 } $. Also, we have that (I use the method Vieta Jumping) :
$ \displaystyle{ \mathbb{S} = \Big \{ \big( a, b \big) \in \mathbb{N} \times \mathbb{N} : \frac{a + 1}{b} + \frac{b + 1}{a} = k \in \mathbb{N} \Big \} = \Big \{ \big( a, b \big) \in \mathbb{N} \times \mathbb{N} : \frac{a + 1}{b} + \frac{b + 1}{a} = 3 \text{  } or \text{  } \frac{a + 1}{b} + \frac{b + 1}{a} = 4 \Big \} } $
Fibonacci sequence : $\displaystyle{f_{0} = 0, f_{1} = 1}$ and $\displaystyle{f_{n+2} = f_{n+1} + f_{n}, \forall n \in \mathbb{N} \cup \{ 0 \} = \{ 0, 1, 2, 3, . . . \}}$
$ \displaystyle{ k = 3 } $ : If $ \displaystyle{ \big( a, b \big) \in \mathbb{S} } $ with $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 3 } $, then $ \displaystyle{ \big( 3 \cdot a - 1 - b, a \big) \in \mathbb{S} } $ so we have $ \displaystyle{ \big( 2, 2 \big), \big( 3 = f_{3} + 1, 2 \big), \big( 6 = f_{5} + 1, 3 \big), \big( 14 = f_{7} + 1, 6 \big), \big( 35 = f_{9} + 1, 14 \big), . . . \in \mathbb{S} } $ (I checked some couples and I found that the request is valid) but I can not porve that $ \displaystyle{ \frac{a + b}{\gcd{\big(a, b \big)^{2}}} } $ is a Fibonacci number.
$ \displaystyle{ k = 4 } $ : If $ \displaystyle{ \big( a, b \big) \in \mathbb{S} } $ with $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 4 } $, then $ \displaystyle{ \big( 4 \cdot a - 1 - b, a \big) \in \mathbb{S} } $ so we have $ \displaystyle{ \big( 1, 1 \big), \big( 2, 1 \big), \big( 6, 2 \big), \big( 21, 6 \big), \big( 77, 21 \big), . . . \in \mathbb{S} } $ (I checked some couples and I found that the request is valid) but also I can not porve that $ \displaystyle{ \frac{a + b}{\gcd{\big(a, b \big)^{2}}} } $ is a Fibonacci number.
I need some help, because I can not solve the problem. So my questions are :
if $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 3 } $ where $ \displaystyle{ a = f_{2 \cdot n - 1} + 1, b = f_{2 \cdot n + 1} + 1 \in \mathbb{N}, \forall n \geqslant 2, \forall n \in \mathbb{N} } $, how I can show that $ \displaystyle{ \frac{a + b}{\gcd{\big( a, b \big)^{2}}} } $ is a Fibonacci number ?
and
if $ \displaystyle{ \frac{a + 1}{b} + \frac{b + 1}{a} = 4} $ where $ \displaystyle{ a, b \in \mathbb{N} } $, how I can show that $ \displaystyle{ \frac{a + b}{\gcd{\big( a, b \big)^{2}}} } $ is a Fibonacci number ?
 A: First of all we assume that $gcd(a,b)=d $ so we may assume that there exist coprime integers $x,y$ such that $a=dx , b=dy$ then from the hypothesis of the problem: $ab|a^2+b^2+a+b 
\rightarrow ab|(a+b)(a+b+1) 
\rightarrow dxy|(x+y)(dx+dy+1)$
because $d$ is coprime to $dx+dy+1$ , $d|x+y$ (first result) and because $x,y$ are coprime it is easy to conclude that $xy$ is coprime to $x+y$ and therefore $xy|dx+dy+1$ (second result) now instead of the first result, assume that there is a positive integer $t$ such that $y=td-x$ then the second result will be $x(td-x)|td^2+1 $.
Now assume a positive integer $k$ such that $td^2+1=ktdx-kx^2$ which is the equation $kx^2-ktdx+(td^2+1)=0$ and by multiplying both sides by $4k$, $4k^2x^2-4k^2tdx+(4ktd^2+4k)=0$ and this is actually $(2kx-ktd)^2=(k^2t^2d^2-4k-4ktd^2)        $.
From this result, the existance of such $x$ is clearly equivalent to these two:
i) $k^2t^2d^2-4k-4ktd^2$(we'll call this $r$) is a perfect square
ii) $2k|\sqrt{r}+ktd$. Now just let's say that we choose $k$ to be odd. Then this condition will be changed to the following two conditions:

*

*$2|\sqrt{r}+ktd$ which is clear because $\sqrt{r}$ has the same parity with $r$ and $r$ has the same parity with $k^2t^2d^2$ so actually $\sqrt{r}$ has the same parity with $k^2t^2d^2$ and thus, $\sqrt{r}+ktd$ is even because $ktd+k^2t^2d^2$ is even, so the first condition is satisfied.

*$k|\sqrt{r}+ktd$ iff $k|\sqrt{r}$ iff $k^2|r$ iff $k^2|4k+4ktd^2$ iff $k|4+4td^2$ but because we've chosen $k$ to be odd, that's equivalent to $k|td^2+1$
So, summing up; if we choose $s$ to be a number that satisfies $td^2+1=ks$, it's enough to prove that $r=4k^2s-k^2t^2d^2$ is a perfect square, which is again equivalent to $4s-t^2d^2$ being a perfect square.
So, for avoiding any confusion here I restate the new problem:
We have positive integers that satisfy $td^2+1=ks$ and $4s-t^2d^2$ is a perfect square and want to prove that $t$ is a fibbonachi number.
Now $k$ doesn't have a role anymore so instead of $td^2+1=ks$, we say that $s|td^2+1$.
Now again assume $u$ to be a number that $4s-t^2d^2=u^2$.
Mixing the two hypotheses, we'll have $u^2+t^2d^2=4s|4td^2+4$.Now because RHS shall be bigger than or equal to the LHS, $4td^2+4\leqslant u^2+t^2d^2$.
Now considering $t$ as our variable,
$t^2d^2-4td^2+(u^2-4)$ is less than or equal to zero.
On the other hand that is actually $4+4d^2-u^2\leqslant(t-2)^2d^2$.
Now two important notes here:

*$d\not=1$ because then the problem's original statement leads to a contradiction so I assume it meant $d\not=1$.

*$u\leqslant2$:
If $u=0$ , then back to the $u^2+t^2d^2|4td^2+4$, $t|t^2d^2|4td^2+4$ and thus $t|4$ but if $t=4$ then $16d^2|16d^2+4$ a contradiction. Otherwise $t=1or2$ which are both fibbonachi numbers.
If $u=1$, back to the definition of $u$, $4s=t^2d^2+1$ which is a contradiction $mod4$. So $u\leqslant2$.
Now back to our inequality, $4+4d^2-u^2\leqslant(t-2)^2d^2$. This is equivalent to $sqrt{\frac{4+4d^2-u^2}{d^2}}+2\leqslant t$
Now by the second note we just mentioned, the LHS is maximized when u=2 , and so clearly $2\leqslant t$ and so $t$ is a fibbonachi number.
In the same way you can consider that $k$ is even and get to the numbers {3,5} and so all of the possible values of $t$ are {1,2,3,5} which are all fibbonachi numbers.

