Baltic Way shortlist

By applying Sophie-Germain identity we can obtain following equality: $$\frac{(2m-1{)}^4+\frac{1}{4}}{(2m{)}^4+\frac{1}{4}}=\frac{((2m-\frac{1}{2}{)}^2+\frac{1}{4})((2m-\frac{3}{2}{)}^2+\frac{1}{4})}{((2m+\frac{1}{2}{)}^2+\frac{1}{4})((2m-\frac{1}{2}{)}^2+\frac{1}{4})}=\frac{(2m-\frac{3}{2}{)}^2+\frac{1}{4}}{(2m+\frac{1}{2}{)}^2+\frac{1}{4}}$$ It is now easy to see that the product, that we want to compute, can be shortened: $$\prod _{m=1}^{2018}\frac{(2m-1{)}^4+\frac{1}{4}}{(2m{)}^4+\frac{1}{4}}=\prod _{m=1}^{2018}\frac{(2m-\frac{3}{2}{)}^2+\frac{1}{4}}{(2m+\frac{1}{2}{)}^2+\frac{1}{4}}=\frac{(2-\frac{3}{2}{)}^2+\frac{1}{4}}{(2\cdot 2018+\frac{1}{2}{)}^2+\frac{1}{4}}=\frac{\frac{1}{2}}{\frac{8073^2+1}{4}}=\frac{2}{8073^2+1}$$