PROBLEM 4. Determine the values of $n \in \mathbb{N}$, for which

$$
I_{n}=\int_{0}^{1} \frac{(1+x)^{n}+(1-x)^{n}}{1+x^{2}} \mathrm{~d} x \in \mathbb{Q}
$$[^0]

## NATIONAL MATHEMATICS OLYMPIAD Local stage - 14.02. 2015 GRADING GUIDE - 12th Grade
## PROBLEM 4.

(2p) By decomposing the rational function $f(x)=\frac{(1+x)^{n}+(1-x)^{n}}{1+x^{2}}$ into partial fractions, we obtain $f(x)=A(x)+\frac{p x+q}{1+x^{2}}$, where $p, q \in \mathbb{Z}$.

(1p) From $(1+x)^{n}+(1-x)^{n}=\left(1+x^{2}\right) \cdot A(x)+p x+q$, we get $\left\{\begin{array}{rrr}(1+i)^{n}+(1-i)^{n} & =p i+q \\ (1-i)^{n}+(1+i)^{n} & =-p i+q\end{array} \Rightarrow\left\{\begin{array}{l}p=0 \\ q=(1-i)^{n}+(1+i)^{n}\end{array}\right.\right.$

(1p) $q=\left(\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\right)^{n}+\left(\sqrt{2}\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)\right)^{n}=2(\sqrt{2})^{n} \cdot \cos \frac{n \pi}{4}$

(1p) We have $I_{n}=\int_{0}^{1} A(x) \mathrm{d} x+\int_{0}^{1} \frac{q}{1+x^{2}} \mathrm{~d} x$,

(1p) where $\int_{0}^{1} A(x) \mathrm{d} x \in \mathbb{Q}$ and $\int_{0}^{1} \frac{q}{1+x^{2}} \mathrm{~d} x=\frac{\pi}{2} \cdot(\sqrt{2})^{n} \cdot \cos \frac{n \pi}{4}$

(1p) Therefore, $I_{n} \in \mathbb{Q} \Leftrightarrow \int_{0}^{1} \frac{q}{1+x^{2}} \mathrm{~d} x \in \mathbb{Q} \Leftrightarrow \frac{\pi}{2} \cdot(\sqrt{2})^{n} \cdot \cos \frac{n \pi}{4} \in \mathbb{Q} \Leftrightarrow \cos \frac{n \pi}{4}=0$, from which we obtain that $n \in\{4 k+2: k \in \mathbb{N}\}$.[^1]


[^0]:    ${ }^{1}$ The actual working time is 3 hours;

    ${ }^{2}$ All problems are mandatory;

    ${ }^{3}$ Each problem is graded from 0 to 7 .

[^1]:    ${ }^{1}$ Each corrector awards an integer number of points;

    ${ }^{2}$ Any other correct solution is graded accordingly.