XVI OM - II - Task 4

Find all prime numbers $ p $ such that $ 4p^2 +1 $ and $ 6p^2 + 1 $ are also prime numbers.
To solve the problem, we will investigate the divisibility of the numbers \( u = 4p^2 + 1 \) and \( v = 6p^2 + 1 \) by \( 5 \). It is known that the remainder of the division of the product of two integers by a natural number is equal to the remainder of the division of the product of their remainders by that number. Based on this, we can easily find the following:
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When \( p \equiv 0 \pmod{5} \), then \( u \equiv 1 \pmod{5} \), \( v \equiv 1 \pmod{5} \);\\
When \( p \equiv 1 \pmod{5} \), then \( u \equiv 0 \pmod{5} \), \( v \equiv 2 \pmod{5} \);\\
When \( p \equiv 2 \pmod{5} \), then \( u \equiv 2 \pmod{5} \), \( v \equiv 0 \pmod{5} \);\\
When \( p \equiv 3 \pmod{5} \), then \( u \equiv 2 \pmod{5} \), \( v \equiv 0 \pmod{5} \);\\
When \( p \equiv 4 \pmod{5} \), then \( u \equiv 0 \pmod{5} \), \( v \equiv 2 \pmod{5} \).\\
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From the above, it follows that the numbers \( u \) and \( v \) can both be prime numbers only when \( p \equiv 0 \pmod{5} \), i.e., when \( p = 5 \), since \( p \) must be a prime number. In this case, \( u = 4 \cdot 5^2 + 1 = 101 \), \( v = 6 \cdot 5^2 + 1 = 151 \), so they are indeed prime numbers.
Therefore, the only solution to the problem is \( p = 5 \).