$\pi\rightarrow e$

The key fact is that $S_a = \sum_{n=-\infty}^\infty \frac1{n^2+a^2} = \frac{\pi}a \coth(\pi a).$ This has a lot of classical proofs, which you can look up (e.g. here ).

Applying this with $a = \frac1{2\pi}$ gives $\begin{aligned} \sum_{n=-\infty}^\infty \frac{4\pi^2}{4\pi^2 n^2 + 1} &= 2\pi^2 \coth(1/2) \\ 2 \sum_{n=-\infty}^\infty \frac1{4\pi^2 n^2 + 1} &= \coth(1/2) \\ 2 \left( 1 + 2\sum_{n=1}^\infty \frac1{4\pi^2n^2 + 1} \right) &= \coth(1/2) \\ \sum_{n=1}^\infty \frac1{4\pi^2 n^2 + 1} &= \frac{\coth(1/2)}4-\frac12. \end{aligned}$

Now $\begin{aligned} \frac{\coth(1/2)}4-\frac12 &= \frac14 \frac{e^{1/2} + e^{-1/2}}{e^{1/2} - e^{-1/2}} -\frac12 \\ &= \frac{e+1}{4(e-1)} - \frac12 \\ &= \frac{e+1-2(e-1)}{4(e-1)} \\ &= \frac{3-e}{4e-4}. \end{aligned}$ So the answer is $3^2+1^2+4^2-1 = \fbox{25}.$

$\begin{aligned} S & = \sum_{n=1}^\infty \frac 1{(2n\pi)^2+1} \\ & = \sum_{n=1}^\infty \frac 1{(1+2n\pi i)(1-2n\pi i)} \\ & = \frac 12 \sum_{n=1}^\infty \left(\frac 1{1+2n\pi i} + \frac 1{1-2n\pi i}\right) \\ & = \frac i{4\pi} \sum_{n=1}^\infty \left(\frac 1{n+ \frac i{2\pi}} - \frac 1{n - \frac i{2\pi}}\right) & \small \color{#3D99F6} \text{By }\psi (1+z) = - \gamma + \sum_{n=1}^\infty \left(\frac 1n - \frac 1{n+z}\right) \\ & = \frac i{4\pi} \left(\psi \left(1-\frac i{2\pi}\right) - \psi \left(1+\frac i{2\pi}\right) \right) & \small \color{#3D99F6} \text{where }\psi (\cdot) \text{ denotes the digamma function.} \\ & = \frac i{4\pi} \left(\pi \cot \frac i2 + 2\pi i \right) & \small \color{#3D99F6} \text{Since }\psi (1-z) = \psi(z) + \pi \cot(\pi z) \\ & = \frac 14 \left(\frac {e^\frac 12 + e^{-\frac 12}}{e^\frac 12 - e^{-\frac 12}} -2 \right) & \small \color{#3D99F6} \text{and }\psi (1+z) = \psi(z) + \frac 1z \\ & = \frac {3-e}{4e-4} \end{aligned}$

Therefore, ${\color{#D61F06} \mathfrak A}^2 + {\color{#20A900} \mathfrak B}^2 + {\color{#3D99F6} \mathfrak C}^2 - 1 = {\color{#D61F06}3}^2 + {\color{#20A900}1}^2 + {\color{#3D99F6}4}^2 - 1 = \boxed {25}$ .