2002 Indonesia MO Problems/Problem 6

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Problem

Find all prime number $p$ such that $4p^2 + 1$ and $6p^2 + 1$ are also prime.

Solution

If $p = 2$, then $4p^2 + 1 = 17$ and $6p^2 + 1 = 25$. Since $25$ is not prime, $p$ can not be $2$. If $p = 5$, then $4p^2 + 1 = 101$ and $6p^2 + 1 = 151$. Both of the numbers are prime, so $p$ can be $5$.

The rest of the prime numbers are congruent to $1$,$3$,$7$, and $9$ modulo $10$, so $p^2$ is congruent to $1$ or $9$ modulo $10$. If $p^2 \equiv 1 \pmod{10}$, then $4p^2 + 1 \equiv 5 \pmod{10}$. If $p^2 \equiv 9 \pmod{10}$, then $6p^2 + 1 \equiv 5 \pmod{10}$. That means if $p$ is congruent to $1$,$3$,$7$, or $9$ modulo $10$, then either $4p^2 + 1$ or $6p^2 + 1$ can be written in the form $10n + 5$.

The only way for $10n + 5$ to equal $5$ is when $p = 1$ or $p = \tfrac{\sqrt{6}}{3}$, which are not prime numbers. Thus, the rest of the primes can not result in $4p^2 + 1$ and $6p^2 + 1$ both prime, so the only solution is $p = \boxed{5}$.

See Also

2002 Indonesia MO (Problems)
Preceded by
Problem 5
1 2 3 4 5 6 7 Followed by
Problem 7
All Indonesia MO Problems and Solutions
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