Find $\frac{m}{n}$ if
$m=(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)$
and
$n=(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)$
I tried to factor them according to wolfram alpha polynomial factorizer any paranthese has a factor(by putting the upper ones $12k-2$ and the others $12k-8$.
A full solution is given at the "Art of Problem solving" here, and uses indeed the Identity of Sophie Germain identity: The Sophie Germain Identity states that $a^4 + 4b^4$ can be factorized as $(a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab)$. Each of the terms is in the form of $x^4 + 324$. Using Sophie-Germain, we get that $x^4 + 4\cdot 3^4 = (x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x)(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x) = (x(x-6) + 18)(x(x+6)+18)$.
$\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]}$
$= \frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}$ Almost all of the terms cancel out! We are left with $\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}$.