\section*{Exercise 1 - 081221}

Give all prime numbers \(p\) for which both \(p+10\) and \(p+14\) are also prime numbers!
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Let \(a \in \{0,1,2\}\) be the remainder of \(p\) when divided by 3.

If \(a=1\), then \(p+14\) is divisible by 3 and since \(p+14>3\), it is not a prime number.

If \(a=2\), then similarly \(p+10\) is divisible by 3 and is not a prime number. Therefore, \(p\) must be divisible by 3, so \(p=3\).

Then \(p+10=13, p+14=17\) and since these are prime numbers, there is exactly one solution, namely \(p=3\).