How can I write Fermat's big theorem and the twin prime conjecture in quantifier notation?

Question

These are the text descriptions of the two theorems, but I'm a little bit lost as to how I would write these in quantifier notation.

(a) There exist infinitely many pairs of prime numbers of form p and p + 2. (Twin prime conjecture)

(b) There is no nonzero integer solution (x,y,z) for the equation x^n+y^n=z^n, where n is any integer bigger than 2. (Fermat's big theorem)

Answer 1

There exists infinitely many prime numbers $p$ such that $p+2$ is also prime:

$$\exists^\infty p\in\mathbb{P} ~~~~ p+2 \in \mathbb{P}$$

For all positive integers $n$ greater than $2$, there does not exist a positive integer triplet that satisfies $x^n + y^n = z^n$:

$$\forall n\in \mathbb{Z}^+, n > 2 ~~~~\not\exists(x, y, z) \in{\mathbb{Z}^+}^3~~~~x^n + y^n = z^n$$

Answer 2

If you don't have fancy quantifiers like $\exists^\infty$, you can think of a) as:

"There is a prime number $p$ such that $p + 2$ is also a prime, and for every prime number $p$ such that $p + 2$ is also a prime, there exists a larger prime number $p'$ such that $p' + 2$ is also a prime"

This will give you infinitely twin primes, since there is at least one, and that means there is a second, larger, one, and that means there is a third, larger, one, etc.

Assuming your domain is the natural numbers, we can express "$p$ is a prime' as:

$\exists x \exists y (p = x * y \land x \not = 1 \land x \not = p)$

If we abbreviate this expression as 'Prime(p)', we get the following for a):

$\exists p (Prime(p) \land Prime(p + 2)) \land \forall p ((Prime(p) \land Prime(p + 2)) \rightarrow \exists p' (p < p' \land Prime(p') \land Prime(p' + 2)))$

(if you don't like the abbrevation:)

$\exists p (\exists x \exists y (p = x * y \land x \not = 1 \land x \not = p) \land \exists x \exists y (p + 2 = x * y \land x \not = 1 \land x \not = p + 2)) \land \forall p ((\exists x \exists y (p = x * y \land x \not = 1 \land x \not = p) \land \exists x \exists y (p + 2 = x * y \land x \not = 1 \land x \not = p + 2)) \rightarrow \exists p' (p < p' \land \exists x \exists y (p' = x * y \land x \not = 1 \land x \not = p') \land \exists x \exists y (p' + 2 = x * y \land x \not = 1 \land x \not = p' + 2)))$

(which is exactly why abbrevations are good ...)

For b), again assuming your domain is natural numbers:

$\forall n (2 < n \rightarrow \neg \exists x \exists y \exists z (x \not = 0 \land y \not = 0 \land z \not = 0 \land x^n + y^n = z^n)$