Czech Republic competition problems

Note that
$$r(a,b) := a + \frac{1}{a} + b + \frac{1}{b} = \frac{(a+b)(ab+1)}{ab}.$$
We put $a = \frac{t}{u}$ and $b = \frac{v}{w}$, where $t, u, v$ and $w$ are positive integers such that both $t$ and $u$ and also $v$ and $w$ are coprime. Then we get $r(a,b) = \frac{(tv+uw)(tw+uv)}{tuvw}$, whence the Diophantine equation

$$tu(v^2 + w^2) + vw(t^2 + u^2) = kptuvw \quad (6)$$

has to be investigated. Now $\gcd(tu, t^2 + u^2) = 1$. Therefore, (6) implies $tu \mid vw$. As we get similarly $vw \mid tu$, too, we infer

$$tu = vw \quad (7)$$

and (6) becomes

$$\frac{(v^2 + t^2)(v^2 + u^2)}{v^2} = t^2 + u^2 + v^2 + w^2 = kptu.$$

Therefore, $p$ has to divide either $v^2 + t^2$ or $v^2 + u^2$. In the case $p \equiv -1 \pmod 4$, i.e. when $-1$ is a quadratic non-residue mod $p$, this means that $p$ divides $v$ (and $t$ or $u$). But since the same argument is valid for $w$ instead of $v$, we have $p \mid v, w$ contradicting the coprimality of $v$ and $w$. Thus the infinitely many primes with $p \equiv -1 \pmod 4$ have no fantastic multiple and part (a) is solved.

For part (b) we choose $v = 1$ and substitute $w = tu$. Thus we are looking for integers $t$ and $u$ such that

$$1 + t^2 + u^2 + t^2u^2 = kptu.$$

Here we choose $t = F_{2l+1}$, $u = F_{2l-1}$ and use the identity $1 + F_{2l+1}^2 = F_{2l+3}F_{2l-1}$ to obtain
$$(1+t^2)(1+u^2) = (1+F_{2l+1}^2)(1+F_{2l-1}^2) = F_{2l+3}F_{2l-1}F_{2l+1}F_{2l-3} = k p F_{2l+1} F_{2l-1},$$
i.e. $F_{2l+3}F_{2l-3} = kp$. Therefore every prime factor of the Fibonacci number $F_{2l+3}$ has a fantastic multiple.

In view of the well-known formula $\text{gcd}(F_a, F_b) = F_{\text{gcd}(a,b)}$ it is clear that $F_a$ and $F_b$ are relatively prime, if $a$ and $b$ are different prime numbers. Hence we know that infinitely many prime numbers have a fantastic multiple, which solves part (b).