Difference between revisions of "2002 Indonesia MO Problems/Problem 6"

Latest revision as of 23:10, 3 August 2018

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 $p = \boxed{5}$ .

@@ Line 12: / Line 12: @@
 ==See Also==
-{{Indonesia MO 7p box
+{{Indonesia MO box
 |year=2002
 |num-b=5
 |num-a=7
+|eight=
 }}
 [[Category:Intermediate Number Theory Problems]]

2002 Indonesia MO (Problems)
Preceded by Problem 5	1 • 2 • 3 • 4 • 5 • 6 • 7	Followed by Problem 7
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Difference between revisions of "2002 Indonesia MO Problems/Problem 6"

Latest revision as of 23:10, 3 August 2018

Problem

Solution

See Also