1987 AIME Problem 14

We can see that $324=4\cdot3^4$ .

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 $x^4 + 4\cdot 3^4 = (x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x)(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x) = (x(x-6) + 18)(x(x+6)+18)$

Rewriting $\begin{aligned} &= \frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]} \\ &= \frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)} \end{aligned}$

Almost all of the terms cancel out. We are left with $\dfrac{58(64)+18}{4(-2)+18} = \dfrac{3730}{10} = \boxed{373}$

Similar solution with @Adhiraj Dutta's

The given expression can be written as:

$\begin{aligned} P & = \prod_{k=0}^4 \frac {(12k+10)^4+324}{(12k+4)^4+324} \\ & = \prod_{k=0}^4 \frac {(12k+10)^4+4 \cdot 3^4}{(12k+4)^4+4\cdot 3^4} & \small \blue{\text{By Sophie Germain identity}} \\ & = \prod_{k=0}^4 \frac {\left((12k+10+3)^2+3^2\right)\left((12k+10-3)^2+3^2\right)} {\left((12k+4+3)^2+3^2\right)\left((12k+4-3)^2+3^2\right)} \\ & = \prod_{k=0}^4 \frac {\left((12k+13)^2+3^2\right)\cancel{\left((12k+7)^2+3^2\right)}} {\cancel{\left((12k+7)^2+3^2\right)}\left((12k+1)^2+3^2\right)} \\ & = \prod_{k=0}^4 \frac {(12k+13)^2+3^2} {(12k+1)^2+3^2} \\ & = \frac {\prod_\red{k=1}^\red 5 (12k+1)^2+3^2} {\prod_{k=0}^4 (12k+1)^2+3^2} \\ & = \frac {(12(5)+1)^2+3^2}{(12(0)+1)^2+3^2} = \frac {3721+9}{1+9} = \boxed{373} \end{aligned}$

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Reference: Sophie Germain identity