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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

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

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

$$tu=vw(7)$$

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 $p\equiv -1(\mathrm{mod}4)$, 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 $p\equiv -1(\mathrm{mod}4)$ 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

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

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 $$(1+t^2)(1+u^2)=(1+F_{2l+1}^2)(1+F_{2l-1}^2)=F_{2l+3}{F}_{2l-1}{F}_{2l+1}{F}_{2l-3}=kp{F}_{2l+1}{F}_{2l-1},$$ 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).