Irish Mathematical Olympiad

If $p \equiv 2 \pmod{3}$ and $x = y = 1$, the number
$$x + y + \frac{p}{x} + \frac{p}{y} = 2(1 + p)$$
is an integer that is divisible by 3, hence for all such primes the required solutions $x, y, n$ exist.

We now show that solutions can only exist if $p \equiv 2 \pmod{3}$.

Suppose
$$x = \frac{a}{b} \quad \text{and} \quad y = \frac{c}{d}$$
with positive integers $a, b, c, d$ that satisfy $\gcd(a, b) = \gcd(c, d) = 1$, is a solution. The given equation is then equivalent to
$$(a^2 + p b^2) c d + (c^2 + p d^2) a b = 3 n a b c d. \quad (1)$$

Case (i) Because $a^3 = a(a^2 + p b^2) - p b (a b)$, any prime factor of $\gcd(a b, a^2 + p b^2)$ must divide $a$. Hence, $p$ is not among these prime factors and such a prime factor must divide $p b^2 = (a^2 + p b^2) - a^2$. This contradicts $\gcd(a, b) = 1$. Hence, $\gcd(a b, a^2 + p b^2) = 1$. Similarly, using $p \nmid c, w_2$ obtain $\gcd(c d, c^2 + p d^2) = 1$.

Equation (1) shows that $a b$ divides $(a^2 + p b^2) c d$. Because $\gcd(a b, a^2 + p b^2) = 1$, Euclid's Lemma implies that $a b \mid c d$. Similarly, $c d$ divides $(c^2 + p d^2) a b$ and, using $\gcd(c d, c^2 + p d^2) = 1$ we obtain $c d \mid a b$. This shows that $a b = c d$.

As $c d \neq 0$ we can now cancel $c d$ in (1) and obtain
$$a^2 + p b^2 + c^2 + p d^2 = 3 n a b.$$
If $p \equiv 1 \pmod{3}$ we obtain that $a^2 + b^2 + c^2 + d^2 \equiv 0 \pmod{3}$, hence at least one of the four numbers $a, b, c, d$ must be divisible by 3. But because $a b = c d$, there is a second of these numbers divisible by 3. But then the sum of the squares of the remaining two numbers must also be divisible by 3, which is only possible if these two numbers are divisible by 3 themselves. But this contradicts $\gcd(a, b) = 1$.

If $p = 3$ we immediately obtain that $a^2 + c^2$ is divisible by 3 and so $a$ and $c$ are divisible by 3, contrary to our assumption $p \nmid a c$.

Case (ii) If $p \mid a$ we can write $a = p \bar{a}$ with some positive integer $\bar{a}$ and we have $p \nmid b$, as $\gcd(a, b) = 1$. Equation (1) is then equivalent to
$$(p \bar{a}^2 + b^2) c d + (c^2 + p d^2) \bar{a} b = 3 n \bar{a} b c d$$
and we have $\gcd(\bar{a}, b) = \gcd(c, d) = 1$. So, we are again in the situation of Case (i), now with $\bar{a}$ replacing $b$ and $c$ replacing $a$. Therefore, we need to have $p \equiv 2 \pmod{3}$.

Case (iii) If $p \mid a$ and $p \mid c$, we can write $a = p \bar{a}$ and $c = p \bar{c}$ with positive integers $\bar{a}, \bar{c}$. Equation (1) is then equivalent to
$$(p \bar{a}^2 + b^2) \bar{c} d + (p \bar{c}^2 + d^2) \bar{a} b = 3 n \bar{a} b \bar{c} d$$
with $\gcd(\bar{a}, b) = \gcd(\bar{c}, d) = 1$. Moreover, because $p \mid a$ and $\gcd(a, b) = 1$ as well as $p \mid c$ and $\gcd(c, d) = 1$, we also have $p \nmid b d$. Hence, we are again in the situation of Case (i) and conclude that we need to have $p \equiv 2 \pmod{3}$ in this case as well.

The final result is that such numbers $x, y, n$ exist for all primes $p$ that satisfy $p \equiv 2 \pmod{3}$ and for no other prime number.