MathB.in - Share Mathematics with LaTeX and Markdown

In the language of math the key is in how we frame the question. For example the following will provide some insight as I will show in a momemt.

\(\exists n \in \mathbb{N}\) such that \(n^3 - 1\) is prime

Not everyone understands the above notation so let me rephrase it more simply. The above translates to "There exists a Natural Number, \(n\), such that \(n^3 - 1\) is prime." Remember a Natural Numer is any positive Integer. In this example assuming that the original assertion that "8 is the only perfect cube to follow a prime" then \(n = 2\) which means the prime number, \(7\), is \(2^3 - 1 = 7\), and \(8\) is just \(2^3 = 8\). Easy enough, but how can we prove that this is the only case...

So we really just need to figure out which values for n in the equation above will give us a prime number, then we have our answer. A prime number is any number which only has 1 and itself as its factors. In other words the only two natural numbers we could possibly multiply together to get 7 is 1 and 7. So we have to start by factoring out the above equation \(n^3 - 1\), if we do that we get:

\[(n-1) \cdot (n^2+n+1)\]

It should be immediately obvious that of these two factors the left-most one is the smaller number, so we know:

\[(n-1) < (n^2+n+1)\]

Since we only care about prime numbers which satisfy the equation we know the left hand term must be equal to 1 and the right hand term most be equal to the entire number. So we can likewise assert the following:

\[1 = n-1\]
\[n^3 - 1 = n^2+n+1\]

Now we can use either equation and solve for n. It doesnt matter which equation you solve they will both give you the same value for n. Solving for n we get:

\[n = 2\]

If we plug that into the original equation, as we said earlier, er get the answer of 7, therefore 7 is the only prime number followed by a perfect cube.

\[2^3 - 1 = 7\]

Sun, 01 Dec 2019 17:55 GMT