Add Me!Close Menu Navigation

The Mysterious Case of the Hidden Evens

Find all positive integers n for which 3n−4, 4n−5, and 5n−3 are all prime numbers.

We’re asked to find all integers n satisfying 3n-4, 4n-5, 5n-3 are all primes.

The hidden trick behind this problem is that 3n-4+ 4n-5+ 5n-3 = 12n - 12 is even. Now, suppose that the sum of three prime integers turned out to be even, let’s enumerate all the ways to add three numbers to find out how this is possible.

    \[\begin{tabular}{c|c|c|c} p_1 & p_2 & p_3 & p_1 + p_2 + p_3\\ \hline \mbox{odd} & \mbox{odd} & \mbox{odd}  & \mbox{odd}\\ \mbox{even} & \mbox{odd} & \mbox{odd} & \mbox{even} \\ \mbox{even} & \mbox{even} & \mbox{odd} & \mbox{odd}\\ \mbox{even} & \mbox{even} & \mbox{even} & \mbox{even}  \end{tabular}\]

So either exactly one of 3n-4, 4n-5, 5n-3 is even or that all of 3n-4, 4n-5, 5n-3 are even.

Let’s look at the case in which each of the primes 3n-4, 4n-5, 5n-3 is even. The only even prime number that I’m aware of is 2, so that means

    \begin{align*} 3n-4 &= 2 \\ 4n-5 &= 2 \\ 5n-3 &= 2 \\ \end{align*}

This obviously has no solutions because 4n-5 = 2 \iff 4n = 7 | n \in \bf Z is impossible, hence this case cannot hold.

Therefore, it must be the case that exactly one of 3n-4, 4n-5, 5n-3 is even.

Since 4n - 5 = 2 was already shown to be impossible, then either

    \begin{align*} 3n-4 &= 2 \mbox{ or}\\ 5n-3 &= 2 \\ \end{align*}

So both n = 2 and/or n = 1.

Let’s first check if n = 2 is satisfiable:

    \begin{align*} 3(2)-4 &= 2 & \mbox{2 is prime} \\ 4(2)-5 &= 3 & \mbox{3 is prime} \\ 5(2)-3 &= 7 & \mbox{7 is prime} \end{align*}

Hence n = 2 is one solution.

Now, let’s check whether n = 1 holds:

    \[3(1) - 4 < 0 \mbox{ cannot be prime}\]

Therefore, only n=2 satisfies 3n-4, 4n-5, 5n-3 are all prime numbers. \square

Posted By Lee

Woosh