Let's determine those twin prime numbers $p$ and $q$ for which $p^{2}-p q+q^{2}$ is also prime. (The prime numbers $p$ and $q$ are twin primes if $|p-q|=2$.

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We can assume that $p$ is the larger prime, because $p^{2}-p q+q^{2}$ is symmetric with respect to $p$ and $q$.

Let $p=r+1, q=r-1 (r>1$ integer $)$.

Substituting, we get

$$
(r+1)^{2}-(r+1)(r-1)+(r-1)^{2}=r^{2}+3>3
$$

due to the condition. $r^{2}+3$ can only be prime if $r$ is not divisible by 3, since $q, r$, and $p$ are three consecutive numbers. A number divisible by 3 can only be prime if it equals 3. Therefore, two possibilities arise: either $p=3 ; q=1$, or $p=5 ; q=3$.

Since 1 is not a prime, the only solution is $p=5 ; q=3$, or by symmetry, $p=3 ; q=5$.

This satisfies the problem's requirement, since

$$
p^{2}-p q+q^{2}=9-15+25=19
$$

which is also a prime.

Based on the work of Gyula Pap (Debrecen, Fazekas M. Gymnasium, 1st year)