Argentine National Olympiad 2016

The minimum value of $p$ is $p=5$. Write $a$ and $b$ as fractions with least common denominator $w$: $a=\frac{u}{w}$, $b=\frac{v}{w}$. In other words, if $a=\frac{u^{\prime }}{w^{\prime }}$, $b=\frac{v^{\prime }}{w^{\prime }}$ is another representation with common denominator $w^{\prime }$, then $w^{\prime }\ge w$. The irreducible representation $s=\frac{m}{n}$ is obtained from $s=\frac{u+v}{w}$ by possible cancellation. Therefore, the prime divisors of $n$ are among the ones of $w$.

We show that $w$ is not divisible by $2$ and $3$, implying that neither is $n$. The condition $a+b=a^2+b^2$ gives $u^2+v^2=w(u+v)$. Suppose that $3$ divides $w$. Then $u^2+v^2$ is a multiple of $3$, and since $x^2\equiv 0,1(\mathrm{mod}3)$ for each integer $x$, it follows that both $u$ and $v$ are divisible by $3$. However, then $3$ is a common divisor of $u$, $v$, and $w$, which contradicts the minimality of $w$. Similarly, suppose that $w$ is even. Then $u^2+v^2$ is even, hence so is $u+v$ ($u^2+v^2$ has the same parity as $u+v$). Hence $u^2+v^2$ is divisible by $4$, and since $x^2\equiv 0,1(\mathrm{mod}4)$ for each integer $x$, both $u$ and $v$ are even. We reach a contradiction with the minimality of $w$ again.

By the above, each prime divisor of $n$ is at least $5$. For an example with $p=5$, let $a=\frac{2}{5}$, $b=\frac{6}{5}$. Then $a+b=\frac{8}{5}$, $a^2+b^2=\frac{4}{25}+\frac{36}{25}=\frac{40}{25}=\frac{8}{5}$. So $a+b=a^2+b^2$ holds, the common value $s$ is not an integer, and its representation $s=\frac{8}{5}$ is irreducible with $p=n=5$.