One nonstandard is worth infinitely many standards

Suppose that M is a nonstandard model of true arithmetic (the set of all first-order sentences in the language of PA that are true in the standard model of arithmetic, ℕ). Now, take any sentence φ(x) with one free variable. Suppose that there’s some nonstandard number k in M such that φ holds of k. Since k is larger than every standard natural, the following infinite set of sentences are all true:

∃x (x > 0 ∧ φ(x))
∃x (x > 1 ∧ φ(x))
∃x (x > 2 ∧ φ(x))

∃x (x > 1000000 ∧ φ(x))

Since these sentences are true in M and M is a model of true arithmetic, these sentences must also be true in the standard model ℕ. So it must be true for every standard natural that there’s a larger standard natural that satisfies φ. In other words, you can guarantee that there are infinitely many standard naturals that satisfy a property φ just by finding a single nonstandard number k that satisfies φ in a model of true arithmetic!

Furthermore, since in ℕ it is true that every standard natural has a larger standard natural satisfying φ, the sentence ∀x ∃y (y > x ∧ φ(y)) is true in ℕ. So this sentence must be true in every model of true arithmetic, including M! This means that just by finding a single nonstandard satisfying φ, you can immediately be sure that there are infinitely many standard numbers AND infinitely many nonstandard numbers (in every nonstandard model of TA!) satisfying φ. This is pretty dramatic!

As an example, consider the twin prime conjecture. We can construct the predicate isPrime(x) in first-order PA with the formula ∀y (∃z (y⋅z = x) → (y=1 ∨ y=x)). Then the predicate isTwinPrime(x) is just: isPrime(x) ∧ isPrime(x+2). Now the twin prime conjecture just says that ∀x ∃y (y > x ∧ isTwinPrime(y)), which is exactly the form we saw in the last paragraph! So to prove the twin prime conjecture, it suffices to demonstrate a single nonstandard twin prime in a model of true arithmetic.