Number Theory Problem Germany 2003 
Prove that there exist infinitely many pairs $(a,b)$ of relatively prime positive integers
  such that $\frac{a^2-5}{b}, \frac{b^2-5}{a}$ are both positive integers.

I saw that this problem came from Germany 2003, but was unable to find a corresponding solution online. I tried performing casework on $a$ and $b\mod 4$, but came up with nothing.
 A: Hint. Show the infinitude of positive integer solutions $(a,b)$ to the divisibility condition $ab\mid a^2+b^2-5$.  In fact, for a positive integer $k$, there exists $(a,b)\in\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}$ such that $$a^2+b^2-5=kab\tag{*}$$
if and only if $k=3$, in which case there are infinitely many choices of $(a,b)$.  When $k=3$, amongst the positive integer solutions $(a,b)$ such that $a\geq b$, the smallest of which is $(a,b)=(4,1)$.  
The idea is the technique known as Vieta jumping.  If you do this correctly, then you will see that all positive integer solutions $(a,b)$ with $a\geq b$ to (*) with $k=3$  are of the form $(a,b)=(x_n,x_{n-1})$ for some positive integer $n$, where $(x_n)_{n=0}^\infty$ are given by $x_0=1$, $x_1=4$, and
$$x_n=3x_{n-1}-x_{n-2}$$
for every integer $n\geq 2$.  Here is a closed form of $\left(x_n\right)_{n=0}^\infty$:
$$x_n=\left(\frac{1+\sqrt5}{2}\right)^{2n+1}+\left(\frac{1-\sqrt5}{2}\right)^{2n+1}=L_{2n+1}$$
for all $n=0,1,2,\ldots$, where $(L_r)_{r=0}^\infty$ is the sequence of Lucas numbers.  The first few terms of $(x_n)_{n=0}^\infty$ are
$$1,4,11,29,76,199,521,1364,3571,9349,24476,\ldots\,.$$  Compare the list above with the answer by Arthur. 
A: Partial answer: After writing a quick program that checks for solutions, I found, among others, the following pairs that work:
$$
4, 11\\
11, 29\\
29, 76\\
76, 199\\
199, 521\\
521, 1364
$$
They seem like a chain of pairs, each constructed from the previous pair in some way.
To see that these do indeed work, note that we have
$$
\frac{4^2 - 5}{11} = 1, \quad \frac{11^2-5}4 = 29\\
\frac{11^2 - 5}{29} = 4, \quad\frac{29^2-5}{11} = 76\\
\frac{29^2-5}{76} = 11, \quad \frac{76^2 - 5}{29} = 199
$$
Hang on a moment. This looks like a really big coincidence. Let's put words on it, and then see if we can't prove that it is true:

Given a pair $a, b$ that fulfills the criteria of the problem, the pair $b, \frac{b^2 - 5}{a}$ also fulfills the criteria of the problem.

We check:
$$
\cfrac{b^2-5}{\frac{b^2-5}a} = a
$$
is clearly an integer. I'm stuck at the other one:
$$
\frac{\left(\frac{b^2-5}a\right)^2 - 5}b
$$
A: This is from Batominovski's anwer, make this CW. Worth remembering
Suppose we have constant integers $V,W$   and nonzero integer variables $x,y$ with the requirement that we always have $\gcd(x,y) = 1.$ Given the two conditions
$$ x \;| \; y^2 + Ty + U \; , \; $$
$$ y \; | \; x^2 + Vx + U \; , \; $$
then
$$ xy \; |  \; \mbox{stuff}  $$
Proof: First, since $x \; | \; x^2 + Vx,$ we get
$$ x \; | \;  x^2 + Vx + y^2 + Ty + U \; , \; $$ or
$$ x \; | \;  x^2 + y^2 + Vx  + Ty + U \; . \; $$
Second, since $y \; | \; y^2 + Ty,$ we get
$$ y \; | \; y^2 + Ty +  x^2 + Vx +  U \; , \; $$ or
$$ y \; | \;  x^2 + y^2 + Vx  + Ty + U \; . \; $$
As $x,y$ are coprime, we reach
$$ xy \; | \;  x^2 + y^2 + Vx  + Ty + U \; . \; $$
To reverse is easier, given that $xy$ divides the thing, ignore the $y$ on the left side and erase the $x$ terms on the right side, we get back to the condition on $x.$ This also repeats the need for matching constant terms. 
So far, it appears that the constant terms must match. I kept the coefficients 1 on $x^2, y^2$ since that is traditional for Vieta questions. This all generalizes to quadratic forms.
