3. Calculate the value of the expression

$$
\frac{\left(3^{4}+4\right) \cdot\left(7^{4}+4\right) \cdot\left(11^{4}+4\right) \cdot \ldots \cdot\left(2015^{4}+4\right) \cdot\left(2019^{4}+4\right)}{\left(1^{4}+4\right) \cdot\left(5^{4}+4\right) \cdot\left(9^{4}+4\right) \cdot \ldots \cdot\left(2013^{4}+4\right) \cdot\left(2017^{4}+4\right)}
$$
3. Calculate the value of the expression

$$
\frac{\left(3^{4}+4\right) \cdot\left(7^{4}+4\right) \cdot\left(11^{4}+4\right) \cdot \ldots \cdot\left(2015^{4}+4\right) \cdot\left(2019^{4}+4\right)}{\left(1^{4}+4\right) \cdot\left(5^{4}+4\right) \cdot\left(9^{4}+4\right) \cdot \ldots \cdot\left(2013^{4}+4\right) \cdot\left(2017^{4}+4\right)}
$$

Answer: $2020^{2}+1=4080401$.

Let's use the formula

$$
\begin{aligned}
a^{4}+4 & =a^{4}+4 a^{2}+4-4 a^{2}=\left(a^{2}+2\right)^{2}-(2 a)^{2}= \\
& =\left(a^{2}-2 a+2\right)\left(a^{2}+2 a+2\right)=\left((a-1)^{2}+1\right)\left((a+1)^{2}+1\right)
\end{aligned}
$$

Now the expression in the numerator of the desired fraction can be written as the product

$$
\left(2^{2}+1\right) \cdot\left(4^{2}+1\right) \cdot\left(6^{2}+1\right) \cdot \ldots \cdot\left(2018^{2}+1\right) \cdot\left(2020^{2}+1\right),
$$

while the expression in the denominator is written as

$$
\left(0^{2}+1\right) \cdot\left(2^{2}+1\right) \cdot\left(4^{2}+1\right) \cdot \ldots \cdot\left(2016^{2}+1\right) \cdot\left(2018^{2}+1\right)
$$

By canceling the common factors in the numerator and the denominator, we arrive at the answer.

## Criteria

+ Correct solution - 7 points.

+ Arithmetic error with the correct idea of the solution - 6 points.

+ The solution contains the idea of factoring $a^{4}+4$, but there is no further progress - 2 points.

- The problem is reduced to some infinite product or sum - this is not considered progress (0 points).