I need closed form for the product $$P_n=(n^2+1^2)(n^2+2^2)...(n^2+(n-1)^2)$$ where $n\in\mathbb{N}$
I tried using the properties of Gamma function as follows: $$P_n= (n+i)(n-i)(n+2i)(n-2i)...(n+(n-1)i)(n-(n-1)i)$$ where $i=\sqrt{-1}$. Rearrangement of the above product gives $$P_n= \left((n+i)(n+2i)...(n+(n-1)i)\right)\left((n-i)(n-2i)...(n-(n-1)i)\right)$$ $$P_n= \left(i^{n-1}\left(\frac{n}{i}+1\right)\left(\frac{n}{i}+2\right)...\left(\frac{n}{i}+(n-1)\right)\right)\left(i^{n-1}\left(\frac{n}{i}-1\right)\left(\frac{n}{i}-2\right)...\left(\frac{n}{i}-(n-1)\right)\right)$$ $$P_n= \left(i^{2n-2}\left(\frac{i}{n}\right)\frac{n}{i}\left(\frac{n}{i}+1\right)\left(\frac{n}{i}+2\right)...\left(\frac{n}{i}+(n-1)\right)\right)\left(\left(\frac{i}{n}\right)\frac{n}{i}\left(\frac{n}{i}-1\right)\left(\frac{n}{i}-2\right)...\left(\frac{n}{i}-(n-1)\right)\right)$$ So we get $$P_n=\frac{i^{2n}}{n^2} \left(\frac{n}{i}\left(\frac{n}{i}+1\right)\left(\frac{n}{i}+2\right)...\left(\frac{n}{i}+(n-1)\right)\right)\left(\frac{n}{i}\left(\frac{n}{i}-1\right)\left(\frac{n}{i}-2\right)...\left(\frac{n}{i}-(n-1)\right)\right) $$ We can write using Gamma representation of Falling and rising factorials $$P_n=\frac{(-1)^{n}}{n^2} \frac{\Gamma\left(\frac{n}{i}+n\right)}{\Gamma\left(\frac{n}{i}\right)}\frac{\Gamma\left(\frac{n}{i}+1\right)}{\Gamma\left(\frac{n}{i}-n+1\right)} $$ Now we use $\Gamma(z)=(1-z)\Gamma(1-z)$ to get $$P_n=\frac{(-1)^{n}}{n^2} \frac{\Gamma\left(\frac{n}{i}+n\right)}{\Gamma\left(\frac{n}{i}-n+1\right)}\left(\frac{n}{i}\right) $$ $$P_n=\left(\frac{(-1)^{n+1} i}{n}\right) \frac{\Gamma\left(\frac{n}{i}+n\right)}{\Gamma\left(\frac{n}{i}-n+1\right)}$$ For $n=1$ the RHS of above formula is $$i \frac{\Gamma\left(\frac{1}{i}+1\right)}{\Gamma\left(\frac{1}{i}\right)}=i.\frac{1}{i}=1=P_1$$ For $n=2$ the RHS of the above formula is $$-\frac{i}{2}\frac{\Gamma(\frac{2}{i}+2)}{\Gamma(\frac{2}{i}-1)}=-\frac{i}{2}\left(\frac{2}{i}+1\right)\left(\frac{2}{i}\right)\left(\frac{2}{i}-1\right) =5=P_2$$ Any help in getting a closed form for the product will be highly appreciated. Thank you.
The product is equal to $\prod_{k=1}^{n-1}\left(k^{2}+n^{2}\right)$.
Using WolframAlpha, you can see that, using the Pochhammer Symbol, it can be written as $(1-in)_{n-1}(1+in)_{n-1}$ which can also be written with the gamma function as $\frac{\sinh(\pi n)\Gamma(n-in)\Gamma(n+in)}{\pi n}$.