Fake proof that the difference of two consecutive cubes is prime

Question

We can write $(x+1)^3 - x^3$ as the difference of two consecutive cubes, then simplify it to $3x^2 +3x+1$, which as the discriminant is less than $0$ we cannot write it as a product of two linear factors without using complex numbers, therefore the quadratic is always prime. Where is the flaw in this argument?

Answer 1

You have successfully proven the following statement: $$\not\exists a, b, c, d \in \mathbb Z: \forall x\in \mathbb Z: (x+1)^3-x^3=(ax+b)(cx+d)\,.$$

You have not proven the following statement: $$\forall x\in \mathbb Z: \not\exists a, b, c, d \in \mathbb Z: (x+1)^3-x^3=(ax+b)(cx+d)\,.$$

The quantifiers are in the wrong order.

Answer 2

A polynomial $a_0 + a_1 x + a_2 x^2 + \ldots + a_n {x_n}^n\ $ is called irreducible over $\mathbb{R}$ if it is not the product of two (non-trivial) polynomials, each with real coefficients.

An alternative way of saying "irreducible over $\mathbb{R}$" is: "Irreducible in $\mathbb{R}[x]$".

[The trivial factorisation of $a_0 + a_1 x + a_2 x^2 + \ldots + a_n {x_n}^n\ $ is $(a_0 + a_1 x + a_2 x^2 + \ldots + a_n {x_n}^n)(1).]$

You seem to think that just because a polynomial is irreducible , that this is enough for the polynomial to spit out a prime number when you feed it an integer. Firstly, you don't give any reasons as to why you think this might be true. Secondly, it isn't true. A simpler example than the one gave in the question is $f(x) = x^2 + 1$, which is irreducible in $\mathbb{R}[x]$ as the discriminant is $<0.$ However, $f(5), f(7), f(8),$ and many more are not prime numbers.

The fact that your polynomial is the general expression of the difference between two consecutive cubes has nothing to do with your actual question.

Edit:

my reasoning as to why the polynomial must output a prime if x is an integer is because the polynomial cannot be broken down into smaller factors (which would be integers as x is an integer)

Ah. Your reasoning is wrong. It is true that a reducible polynomial will spit out a composite (non-prime) integer for almost every input $x$ (except maybe some of the inputs that makes one of the factors equal to $1$). But this doesn't mean (and has nothing to do with) an irreducible polynomial will always spit out a prime number for every $x.$

Remember: A reducible polynomial is by definition a product of two polynomials, like: $x^2-1$ is reducible because $x^2-1=(x-1)(x+1).$ For almost all integer values of $x$ that you pass into this expression - other than $x=-1,0,1$ - the output will be a composite number, since if both factors are not $1$ then you will get a composite number by definition. There is no obvious comparative statement to be made or conclusion to be drawn if the polynomial is irreducible.

If you want to look further into whether or not sets like, $\{ x^2 + 1: x\in\mathbb{N}\}$ have infinitely many primes, then just be aware that this is an open problem known as Landau's fourth problem. If you are interested in this stuff then start with Dirichlet's theorem, which is known to be true, although even the proof of this is not easy.