Using factorization to prove that some numbers are not prime

There was some discussion on my first-year Pure Maths (FPM) Piazza forum recently of one of my quiz questions about prime numbers.

The main question was how to prove that statement (c) is false. The students' response correctly noted that we can factorize: \(p^3+8 = (p+2)(p^2-2p+4)\), and so \(p^3+8\) can't be prime.

Let's have a more careful look at this. Note that \(p\) is a prime number, and in particular \(p\) is an integer and \(p\geq 2\). (This is part of our definition of prime number. For example, we don't allow numbers like \(1\) or \(-7\) here.)

Set \(m=p^3+8\). We can see from the factorization above that \(p+2\) is a divisor of \(m\), and we definitely have \(1 < p+2 < m\) here. So \(m\) has a positive divisor, \(p+2\), which is neither equal to \(1\) nor \(m\). Thus \(m\) is not prime.

Are there any integers \(p\) at all such that \(p^3+8\) is a prime number? The above factorization suggests this is unlikely, but if \(p={-1}\), then \(p^3+8 = 7\) and \(7\) is prime. So where does the factorization proof above break down in this case? The answer is that when \(p={-1}\), then \(p+2=1\), and so the factorization \(p^3+8 = (p+2)(p^2-2p+4)\) is no longer helpful: it just says that \(1 \times 7 = 7\).

Are there any other examples? Well, if \(p+2 < 1\) part of the above argument still breaks down, where we claim that \(1 < p+2 < m\). On the other hand, if \(m<2\) then \(m\) doesn't satisfy our definition of prime number for that reason instead. This eliminates all integers \(p<{-1}\), because the function \(p \mapsto p^3+8\) is strictly increasing, and so if \(p \leq {-2}\) then \(p^3+8 \leq 0\). So \(-1\) is the only example here.

Exercise: Could there be an integer \(p\) such that \(-(p^3+8)\) is a prime number? If so, how many examples are there?

Hint: This time we can write \(-(p^3+8) = -(p+2)(p^2-2p+4)\).

On Piazza, I posted the following follow-up to the students' solution to the original problem:

The students' response is correct as usual!

Factorization of your expression for \(m\) is a "direct-from-definitions" way to show that an integer \(m>1\) is not prime, as long as you are sure that this gives you at least one divisor \(k\) with \(1<k<m\). But you do need to watch out for the possibility that you have written \(m=m \times 1\) or \(m=(-m) \times (-1)\) in disguise.

For example \(x^3+1=(x+1)(x^2-x+1)\) is never prime for integers \(x>1\) because it has a positive divisor \(x+1\), and we certainly have \(1 < x+1 < x^3+1\) for such \(x\).

However, this argument  goes wrong when \(x=1\), because in that case \(x+1=x^3+1\), and so the proof breaks down. And, indeed, not only does our proof break down (which would be inconclusive in itself), but in fact \(1^3+1=2\), and \(2\) is prime.

Similarly, for positive integers \(a\) and \(b\), \(a^5+b^5\) is not usually prime, because

                     \(a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4)\,,\)

and we would expect that \(1<a+b<a^5+b^5\).

As before, there is exactly one exception to this, which in this case is when \(a=b=1\).

Best wishes,
Dr Feinstein