4. Determine all ordered triples $(a, b, c)$ of real numbers such that the function $f: \mathbb{R} \rightarrow \mathbb{R}$, defined by

$$
f(x)=\sin \left(a x^{2}+b x+c\right)
$$

is even. (A function $f$ is said to be even if $f(x)=$ $f(-x)$ for all $x$.


4. For the observed function to be even, for every real number $x$ it must be satisfied that $\sin \left(a x^{2}+b x+c\right)=\sin \left(a x^{2}-b x+c\right)$, i.e.

$$
0=\sin \left(a x^{2}+b x+c\right)-\sin \left(a x^{2}-b x+c\right)=\sin (b x) \cos \left(a x^{2}+c\right)
$$

For $b=0$ the equality is obviously satisfied, regardless of the values of $a$ and $c$. Now assume $b \neq 0$. Consider the interval $\left(0,\left|\frac{\pi}{b}\right|\right)$. For any $x$ in this interval, we have $\sin (b x) \neq 0$, so it must hold that $\cos \left(a x^{2}+c\right)=0$, i.e., for every $x$ in this interval, it must be satisfied that $a x^{2}+c=\frac{\pi}{2}+k \pi$ for some $k \in \mathbb{Z}$. In the case $a=0$, it must hold that $c=\frac{\pi}{2}+k \pi$, and we see that for any $c$ of this form, the cosine is indeed zero, so the conditions of the problem are satisfied. Now assume $a \neq 0$. Then, as $x$ passes through the interval $\left(0,\left|\frac{\pi}{b}\right|\right)$, $a x^{2}+c$ passes through the interval $\left(c, \frac{a \pi^{2}}{b^{2}}+c\right)$, and it is clearly impossible for all numbers in this interval to be of the form $\frac{\pi}{2}+k \pi$ for some $k \in \mathbb{Z}$ (there are only finitely many such numbers in the given interval); thus, in this case, there is no triplet $(a, b, c)$ that satisfies the conditions of the problem.

Therefore, the answer is: the sought triplets are all triplets of the form $(a, 0, c)$ (where $a$ and $c$ are arbitrary real numbers), as well as all triplets of the form $\left(0, b, \frac{\pi}{2}+k \pi\right)$ (where $b$ is an arbitrary real number and $k$ is an arbitrary integer).