It seems more natural to consider the problem as finding the primes that don't divide and those which divide fantastic numbers. It is essentially a rephrasing of the original problem, but then it almost automatically leads to a solution for part (a). Write $a=\frac{p}{q}$, $b=\frac{r}{s}$, where $\gcd(p,q) = \gcd(r,s) = 1$ (Note that the $p$ here and in the problem aren't the same). Expanding, the original equation becomes $rs(p^2+q^2) + pq(r^2+s^2) = npqrs$.We notice that since $p$ divides the whole equation, $p | rs$. Similarly $q|rs$, so coprimality guarantees $pq | rs$. This can be done the other way round too, so $pq=rs$. So now note that $p^2 + q^2 + r^2 + s^2 = p^2 + q^2 + \frac{p^2q^2}{r^2}+r^2 = \frac{(r^2+q^2)(p^2+r^2)}{r^2} = npq$, which will be our guiding equation for the rest of the problem. This means that all prime factors of $n$ should divide the sum of squares on the left, which, by quadratic reciprocity, is always false for all primes $ \equiv -1 \pmod 4$. For the second part, we characterise all $n$ for which $r=1$ works, so that we can reduce one degree of freedom. we need to only characterise those pairs $(p,q)$ for which $pq | p^2 + q^2 + 1$. However it is well known that the only cases in which this arises is when the quotient is $3$. Thus we need to solve the equation $p^2 + q^2 + 1 = 3 p q$ over the natural numbers, for a family of infinite solutions. To do this, we seek to build solutions from previously found solutions. Suppose we have a solution $(p,q)$, with $p<q$. We need to find a solution $(q,r)$. For this we must have $p^2 + q^2 - 3pq = q^2 + r^2 - 3qr$, which is equivalent to $r = 3q-p$. The base solution is $(1,2)$. So the recurrence has the characteristic equation $x^2-3x+1=0$, which has the roots as the squares of the roots of $x^2-x-1=0$. So by induction, we have one particular family of solutions as $(F_{2j-1}, F_{2j+1})$. So we have $n=3+F_{2j-1}F_{2j+1}$. Using Binet's formula, we know this equals $F_{2j-3}F_{2j+3}$. So all factors of $n$ divide the odd-indexed Fibonacci numbers. But since for Fibonacci numbers, we have $\gcd (F_i, F_j) = F_{\gcd (i,j)}$, we are done.

Well done @above