Infiniteness of non-twin primes. Well, we all know the twin prime conjecture.
There are infinitely many primes $p$, such that $p+2$ is also prime.
Well, I actually got asked in a discrete mathematics course, to prove that there are infinitely many primes $p$ such that $p + 2$ is NOT prime.
Let $p\gt 3$ be prime. If $p+2$ is not prime, we are happy. If $p+2$ is prime, then $(p+2)+2$ is not, since one of $x,x+2,x+4$ is divisible by $3$.

Added: Dolda2000 noted that a more interesting question is whether there are infinitely many primes that are not members of a twin pair. For this we can use the fact that there are infinitely many primes of the form $15k\pm 7$. If $p$ is such a prime, then one of $p-2$ or $p+2$ is divisible by $3$, and the other is divisible by $5$, so if $p\gt 7$ then neither $p-2$ nor $p+2$ is prime.
"I actually got asked in a discrete mathematics course, to prove that there are infinitely many primes p such that p+2 is NOT prime". (I insert the question here to prevent confusing if it was modified.)
I use only that there are infinitely many prime numbers. (An elementary proof of Euclid.)
If $p&gt;3$ ($p$ is prime number) then denote
$$
P_{-1}:=\{p|p=3k-1,\text{for some }k\}
$$
and
$$
P_{1}:=\{p|p=3k+1,\text{for some }k\}.
$$
(Every prime number, greater than $3$, has remainder $\pm 1$ dividing by $3$.)

So at least one of them contains infinitely many prime numbers.

(a) If $P_{1}$ contains finitely many prime numbers then the statement is true, because $P_{-1}$ contains infinitely many prime numbers so if $k$ is large enough then $p+2=(3k-1)+2\in P_{1}$ not a prime number.

(b) If $P_{1}$ contains infinitely many prime numbers then the statement is true, because taking them $3|(3k+1)+2$.