## Subject 1.

Determine the prime numbers $p$ and $q$ knowing that $\frac{p}{q}=\frac{a}{b}+\frac{b}{a}$, where $a$ and $b$ are non-zero natural numbers.

Prof. Lucian Petrescu, Tulcea

| Details of solution | Associated grading |
| :---: | :---: |
| If $a=b$, it follows that $\frac{p}{q}=2$, hence $q=1$, a contradiction. This means that one of the fractions $\frac{a}{b}$ and $\frac{b}{a}$ is greater than 1, so $\frac{p}{q}>1$, hence $p>q$. | 1p |
| We can consider $(a ; b)=1$, then $\frac{p}{q}=\frac{a^{2}+b^{2}}{a b}$. The fraction $\frac{a^{2}+b^{2}}{a b}$ is irreducible, so $q=a b$ and $p=a^{2}+b^{2}$. We deduce that $a=1$ or $b=1$. | $3 \mathbf{p}$ |
| If, for example, $a=1$, we get $q=b$ and $p=b^{2}+1$. Since $p$ and $q$ have different parities and $p>q$, it follows that $q=b=2$ and $p=5$. Indeed, $\frac{5}{2}=\frac{1}{2}+\frac{2}{1}$. | $3 \mathbf{p}$ |
1. Determine the prime numbers $p$ and $q$ knowing that $\frac{p}{q}=\frac{a}{b}+\frac{b}{a}$, where $a$ and $b$ are non-zero natural numbers.