Jake Lai's Problem

$\left[ \prod _{ p=6k+1 \text{ is prime, } p>3 } (1+p^{ -s }) \right] \left[ \prod _{ p=6k-1 \text{ is prime, }p>3 } (1-p^{ -s }) \right] = \prod _{ p \text{ is prime, }p>3 } (1+X(p)p^{ -s })$ Where $X(n)=\left(\frac{n}{3}\right)$ which is the Legendre symbol.

Since $1+X(p)p^{ -s }=\frac { X\left( p \right) +p^{ s } }{ p^{ s } } =\frac { X\left( p \right) }{ p^{ s } } +1=\frac { \frac { 1 }{ p^{ 2s } } -1 }{ \frac { X\left( p \right) }{ p^{ s } } -1 } =\frac { 1-\frac { 1 }{ p^{ 2s } } }{ 1-\frac { X\left( p \right) }{ p^{ s } } }$

$\prod _{ p \text{ is prime, }p>3 } (1+X(p)p^{ -s }) = \frac { \prod _{ p \text{ is prime, }p>3 } \left( 1-\frac { 1 }{ p^{ 2s } } \right) }{ \prod _{ p \text{ is prime, }p>3 } \left( 1-\frac { X\left( p \right) }{ p^{ s } } \right) }$

Now, through Euler's product, $\prod _{ p \text{ is prime} } \left( 1-\frac { 1 }{ p^{ 2s } } \right) =\frac { 1 }{ \zeta (2s) }$ and $\prod _{ p \text{ is prime} } \left( 1-\frac { X\left( p \right) }{ p^{ s } } \right) =\frac { 1 }{ \sum _{ n=1 }^{ \infty } \frac { X\left( n \right) }{ n^{ s } } }$ , where $\zeta (s)$ is the Riemann zeta function.

So, $\frac { \prod _{ p \text{ is prime, }p>3 } \left( 1-\frac { 1 }{ p^{ 2s } } \right) }{ \prod _{ p \text{ is prime, }p>3 } \left( 1-\frac { X\left( p \right) }{ p^{ s } } \right) } = \frac { 6^{ 2s } }{ \left( 6^{ 2s }-2^{ 2s }-3^{ 2s }+1 \right) \zeta (2s) } \left( 1+\frac { 1 }{ 2^{ s } } \right) \sum _{ n=1 }^{ \infty } \frac { X\left( n \right) }{ n^{ s } }$

Since $X\left( n \right) =\begin{cases} 0\text{ if }n\equiv 0\quad \text{mod}(3) \\ -1\text{ if }n\equiv -1\quad \text{mod}(3) \\ 1\text{ if }n\equiv 1\quad \text{mod}(3) \end{cases}$

$\sum _{ n=1 }^{ \infty } \frac { X\left( n \right) }{ n^{ s } } =\sum _{ n=0 }^{ \infty } \left( \frac { 1 }{ \left( 3n+1 \right) ^{ s } } -\frac { 1 }{ \left( 3n+2 \right) ^{ s } } \right) \\$

Therefore, $\frac { \prod _{ p \text{ is prime, }p>3 } \left( 1-\frac { 1 }{ p^{ 2s } } \right) }{ \prod _{ p \text{ is prime, }p>3 } \left( 1-\frac { X\left( p \right) }{ p^{ s } } \right) }= \frac { 6^{ 2s } }{ \left( 6^{ 2s }-2^{ 2s }-3^{ 2s }+1 \right) \zeta (2s) } \left( 1+\frac { 1 }{ 2^{ s } } \right) \sum _{ n=0 }^{ \infty } \left( \frac { 1 }{ \left( 3n+1 \right) ^{ s } } -\frac { 1 }{ \left( 3n+2 \right) ^{ s } } \right)$

Substituting $s=1$ , we get: $\frac{54}{4\pi ^2} \sum _{ n=0 }^{ \infty } \left( \frac { 1 }{ 3n+1 } -\frac { 1 }{ 3n+2 } \right) =\frac { 54 }{ 4\pi ^{ 2 } } \sum _{ n=0 }^{ \infty } \int _{ 0 }^{ 1 }{ { x }^{ 3n }+ } { x }^{ 3n+1 }dx$

$=\frac { 54 }{ 4\pi ^{ 2 } } \frac { \pi }{ 3\sqrt { 3 } } =\frac{3\sqrt{3}}{2\pi }=A$

And hence $\left\lfloor 100000A \right\rfloor =\boxed{82699}$