Proving that $\frac{\pi ^2}{p}\neq \sum_{n=1}^{\infty }\frac{1}{a_{n}^2}$ we have many formula for $\pi ^2$ as follow $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}....$$
and $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}....$$ and for $\frac{\pi ^2}{9}$, $\frac{\pi ^2}{12}$, $\frac{\pi ^2}{16}$,$\frac{\pi ^2}{24}$.. and so on. If you see the number in denominator is not prime.Now we want to prove that
  $$\frac{\pi ^2}{p}\neq \sum_{n=1}^{\infty }\frac{1}{a_{n}^2}$$ Where $p$ is a prime number and $a_{n}$ a positive integer number

I ran the greedy algorithm from the previous question for the first 1000 primes with the added caveat that the terms in the sequence have to be distinct. It appears that your statement is only true for the primes 2, 3, 5, 11, 13 and false for the rest! I've only used 40 terms in the algorithm for each of the primes. For those that work, the accuracy is quite good (hundreds of digits for the larger primes). For example if $A_p$ stands for the terms for prime p then
$$A_7 = \{1, 2, 3, 5, 11, 42, 991, 59880, 21165404, 81090759408,\ldots \}$$
and if $S_p(n)$ stands for the partial sum we have
$$S_7(10) = \frac{9409086091684410250487330038662170074145032522104397}{66733781786148561048500788381308231
   67971682291718400} $$
with
$$\frac{\pi^2}{7}- S_7(10) \approx 3.0292 \cdot 10^{-34}$$
EDIT: I ran the algorithm again but this time with the integers from 1 to 100 because I suspected the problem is not just with the primes themselves. The algorithm failed to converge within 15 terms for the numbers
$$E = \{1, 2, 3, 4, 5, 10, 11, 12, 13, 14, 15\}$$
The algorithm is way off for these integers. I'm willing to conjecture that for every integer $k\notin E$ there exists a sequence of positive integers $a_{n,k}$ such that
$$\frac{\pi^2}{k} = \sum_{n = 1}^{\infty} \frac{1}{a_{n,k}^2}$$
