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Tuesday, March 23, 2010

Solution: Happy Saint Math Trick's Day

Original problem: http://dharmath.blogspot.com/2010/03/happy-saint-math-tricks-day.html

This problem is inspired by this comic.

Prove that the trick in the comic always works. In other words, if $p$ is a prime number greater than 3, prove that $p^2 + 14 \equiv 3 \mod 12$

Solution



$p^2 + 14 \equiv 3 \mod 12$ means that $p^2+11$ is divisible by 12, which means that $p^2+11-12 = p^2-1$ is divisible by 12.

If $p$ is a prime number greater than 3, then $p$ is odd. This means that $p$ divided by 4 has remainder either 1 or 3. In both cases, $p^2$ divided by 4 has remainder 1, which means $p^2-1$ is divisible by 4.

If $p$ is a prime number greater than 3, then $p$ divided by 3 has remainder either 1 or 2. In both cases, $p^2$ divided by 3 has remainder 1, which again means $p^2-1$ is divisible by 3.

Since $p^2-1$ is divisible by both 3 and 4, it is divisible by 12.

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