We can prove that every prime that equals 0, 1, 4 (mod 5) are expressible in the form. Any prime dividing the form $a^2 + a - 1$ is 0, 1, 4 (mod 5). Notice that $1^2 + 1 - 1 = 1 , 2^2 + 2 - 1 = 5$, we don't need to consider on the number of numbers in the denominator or the numerator since we can multiply or divide by 1. Denote all the primes that are 1, 4 (mod 5) by $p_1 , p_2 , p_3 ...$ in the increasing order. Let's prove by induction. Suppose that $p_1 , ..., p_{n-1}$ are expressible. Let $p=p_n$ There is a solution to $x^2 + x - 1 \equiv 0 (mod p)$, since $4(x^2 + x - 1) = (2x+1)^2 - 5$ and 5 is quadratic residue mod p because $p \equiv 1, 4 (mod 5)$. In particular, we have a solution that $1< 2x+1 <p$. With this inequality, we can prove that $x^2 + x - 1 < p^2$ which means that there are no prime divisor of $x^2 + x - 1$ bigger than p. So $x^2 + x - 1$ is divisible by $p$ once, and some other primes. The 'some other primes' are from $p_1 , ..., p_{n-1}$ from the induction hypothesis, so dividing by them appropriate times, multiplying or dividing by 1 few times, we get the desired form.

Does anybody have other solutions?

jaydoubleuel wrote: We can prove that every prime that equals 0, 1, 4 (mod 5) are expressible in the form. Any prime dividing the form $a^2 + a - 1$ is 0, 1, 4 (mod 5). Notice that $1^2 + 1 - 1 = 1 , 2^2 + 2 - 1 = 5$, we don't need to consider on the number of numbers in the denominator or the numerator since we can multiply or divide by 1. Denote all the primes that are 1, 4 (mod 5) by $p_1 , p_2 , p_3 ...$ in the increasing order. Let's prove by induction. Suppose that $p_1 , ..., p_{n-1}$ are expressible. Let $p=p_n$ There is a solution to $x^2 + x - 1 \equiv 0 (mod p)$, since $4(x^2 + x - 1) = (2x+1)^2 - 5$ and 5 is quadratic residue mod p because $p \equiv 1, 4 (mod 5)$. In particular, we have a solution that $1< 2x+1 <p$. With this inequality, we can prove that $x^2 + x - 1 < p^2$ which means that there are no prime divisor of $x^2 + x - 1$ bigger than p. So $x^2 + x - 1$ is divisible by $p$ once, and some other primes. The 'some other primes' are from $p_1 , ..., p_{n-1}$ from the induction hypothesis, so dividing by them appropriate times, multiplying or dividing by 1 few times, we get the desired form. "Any prime dividing the form $a^2 + a - 1$ is 0, 1, 4 (mod 5)" --> If you take $a=9$ then $a^2+a-1=99$, and 3 is dividing 99.

parallelines wrote: "Any prime dividing the form $a^2 + a - 1$ is 0, 1, 4 (mod 5)" --> If you take $a=9$ then $a^2+a-1=99$, and 3 is dividing 99. Well by instance $a^2+a-1 = 89$ when $a=9$... If $a^2+a-1 \equiv 1 \pmod{p}$ we easily get by the quadratic formula that $\left(\dfrac{5}{p}\right) = 1$ which by the quadratic reciprocity law implies $\left(\dfrac{p}{5}\right) = 1$ and it is easy to check the only quadratic residues $\pmod{5}$ are $0,1,4$.