Calculate the value of

$$
\frac{2014^{4}+4 \times 2013^{4}}{2013^{2}+4027^{2}}-\frac{2012^{4}+4 \times 2013^{4}}{2013^{2}+4025^{2}}
$$
For better readability, let $n=2013$. Our expression can be rewritten as

$$
\frac{(n+1)^{4}+4 n^{4}}{n^{2}+(2 n+1)^{2}}-\frac{(n-1)^{4}+4 n^{4}}{n^{2}+(2 n-1)^{2}}=\frac{5 n^{4}+4 n^{3}+6 n^{2}+4 n+1}{5 n^{2}+4 n+1}-\frac{5 n^{4}-4 n^{3}+6 n^{2}-4 n+1}{5 n^{2}-4 n+1}
$$

To simplify (because we always need to simplify!), let's try to factorize the numerator of each fraction by its denominator. For example, we want to write $5 n^{4}+4 n^{3}+6 n^{2}+4 n+1 = (5 n^{2}+4 n+1)(\ldots)$. In the parenthesis $(\ldots)$, we can only have terms in $n^{2}$, $n$, or $n^{0}$. Indeed, if we had a power of $n$ greater than 3, the expansion would give a power greater than 5, which is not present in the numerator. After some trial and error, we find that $5 n^{4}+4 n^{3}+6 n^{2}+4 n+1 = (5 n^{2}+4 n+1)(n^{2}+1)$ and similarly that $5 n^{4}-4 n^{3}+6 n^{2}-4 n+1 = (5 n^{2}-4 n+1)(n^{2}+1)$, hence the answer we are looking for is zero.

Remark: To find the parenthesis ( ... ) systematically, we can use the concept of polynomial division. This involves performing a Euclidean division exactly as we learn in elementary school, but replacing numbers with polynomials, i.e., expressions of the form $\sum a_{i} x^{i}$. Here, this gives:

![](https://cdn.mathpix.com/cropped/2024_05_10_76fb8a4d30ed0b9c3f11g-140.jpg?height=405&width=690&top_left_y=1375&top_left_x=731)