3. Let $f(x)=x^{2}-p x+q$. It turned out that $f(p+q)=0$ and $f(p-q)=0$. Find $p$ and $q$.
Answer: All pairs of the form ( $m, 0$), where $m$ is any number, and the pair $(0,-1)$.
Solution. First case. The numbers $p+q$ and $p-q$ are equal. Then $q=0$. The roots of this quadratic trinomial are $p$ and 0. Therefore, all pairs ( $m, 0$ ), where $m$ is any number, are suitable.

Second case. The numbers $p+q$ and $p-q$ are different. Then, by Vieta's theorem, $\mathrm{p}+\mathrm{q}+\mathrm{p}-\mathrm{q}=\mathrm{p},(\mathrm{p}+\mathrm{q})(\mathrm{p}-\mathrm{q})=\mathrm{q}$. From the first equation $\mathrm{p}=0$. From the second $-\mathrm{q}^{2}=\mathrm{q} . \mathrm{q}=-1, \mathrm{q}=0$. The pair $(0,0)$ does not fit this case. The pair ( $0,-$ 1) fits.

Criteria. The case of coinciding roots is not considered: 3 points.