3. Find the value of $f(2019)$, given that $f(x)$ simultaneously satisfies the following three conditions:
1) $f(x)>0$ for any $x>0$
2) $f(1)=1$;
3) $f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}$ for any $a, b \in \mathbb{R}$.
Solution. In the identity given in the problem

$$
f(a+b) \cdot(f(a)+f(b))=2 f(a) \cdot f(b)+a^{2}+b^{2}
$$

let $a=1, b=0$. Then $f(1) \cdot(f(1)+f(0))=2 f(1) \cdot f(0)+1$. Since $f(1)=1$, we find

$$
f(0)=0
$$

Next, by setting $b=-a$ in (1), we get, taking (2) into account, that

$$
f(a) \cdot f(-a)=-a^{2}
$$

Finally, when $b=0$, identity (1) (taking (2) into account) becomes $f(a) \cdot f(a)=a^{2}$.

Thus, it is necessary that $f(a)=a$ for $a>0$, since by condition $f(x)>0$ for $x>0$. Furthermore, according to (3), $f(a)=a$ for $a<0$ as well. Ultimately, $f(x)=x$ for any $x \in \mathbb{R}$. It is easy to verify that such an $f(x)$ indeed satisfies conditions 1), 2), 3) from the problem statement. Therefore, $f(x)=x$.

Answer: 2019.