A very prime product

rewrite this as $\prod_{p} \left(1-\dfrac{1}{p^4}\right) \prod_p\left(1-\dfrac{1}{p^2}\right)^{-2}=\dfrac{\zeta^2(2)}{\zeta(4)} = \dfrac{5}{2}$ where the prime product defination of the riemann zeta function is used

From Euler's product , we have:

$\begin{cases} \displaystyle \prod_p \left(1-p^{-s} \right)^{-1} = \prod_p \frac {p^s}{p^s - 1} = \zeta (s) \\ \displaystyle \prod_p \left(1+p^{-s} \right)^{-1} = \prod_p \frac {p^s}{p^s + 1} = \frac {\zeta (2s)}{\zeta (s)} \end{cases}$ , where $\zeta(\cdot)$ denotes the Riemann zeta function .

Then we have $\displaystyle \prod_p \frac {p^2+1}{p^2-1} = \prod_p \frac {\left(1-p^{-2} \right)^{-1}}{\left(1+p^{-2} \right)^{-1}} = \frac {\zeta^2 (2)}{\zeta(4)} = \frac {\left(\frac {\pi^2}6\right)^2}{\frac {\pi^4}{90}} = \frac 52$ . Therefore, $a+b = 5 + 2 = \boxed 7$ .

This is not a solution, but a simple script I used to find a good guess for the answer before actually solving the problem.

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from fractions import Fraction


def prime_gen():

    def candidate_gen():
        c = 5
        while True:
            yield c
            c += 2
            yield c
            c += 4

    primes = [2, 3]
    yield 2
    yield 3
    for c in candidate_gen():
        for p in primes:
            if p ** 2 > c:
                primes.append(c)
                yield c
                break
            if c % p == 0:
                break


if __name__ == '__main__':
    prod = Fraction(1, 1)
    for p in prime_gen():
        prod *= Fraction(p ** 2 + 1, p ** 2 - 1)
        print(prod.__float__())