Do I have a prime twin?

Brachylog, 9 8 bytes

ṗ∧4√;?+ṗ

Try it online!

Explanation

ṗ           Input is prime
 ∧          And
  4√        A number whose square is 4 (i.e. -2 or 2)
    ;?+     That number + the Input…
       ṗ    …is prime

Jelly, 10 9 bytes

%6+_2_ÆP⁺

Try it online!

Background

With the exception of (3, 5), all twin prime pairs are of the form (6k - 1, 6k + 1).

Since (6k - 1) + (6k - 1) % 6 - 3 = 6k - 1 + 5 - 3 = 6k + 1 and
(6k + 1) + (6k + 1) % 6 - 3 = 6k + 1 + 1 - 3 = 6k - 1, given an input n > 3, it is sufficient to check whether n and n + n % 6 - 3 are both prime.

This formula happens to work for n = 3 as well, as 3 + 3 % 6 - 3 = 3 is prime and 3 is a twin prime.

How it works

%6+_2_ÆP⁺  Main link. Argument: n

%6         Compute n%6.
  +        Add n to the result.
   _2      Subtract 2.
     _ÆP   Subtract 1 if n is prime, 0 if not.
           If n is not a prime, since (n + n%6 - 2) is always even, this can only
           yield a prime if (n + n%6 - 2 = 2), which happens only when n = 2.
        ⁺  Call ÆP again, testing the result for primality.

Python 3, 53 bytes

lambda n:sum((n+n%6-3)*n%k<1for k in range(2,4*n))==2

Try it online!

Background

All integers take one of the following forms, with integer k: 6k - 3, 6k - 2, 6k - 1, 6k, 6k + 1, 6k + 2.

Since 6k - 2, 6k, and 6k + 2 are all even, and since 6k - 3 is divisible by 3, all primes except 2 and 3 must be of the form 6k - 1 or 6k + 1. Since the difference of a twin prime pair is 2, with the exception of (3, 5), all twin prime pairs are of the form (6k - 1, 6k + 1).

Let n be of the form 6k ± 1.

• If n = 6k -1, then n + n%6 - 3 = 6k - 1 + (6k - 1)%6 - 3 = 6k - 1 + 5 - 3 = 6k + 1.

• If n = 6k + 1, then n + n%6 - 3 = 6k + 1 + (6k + 1)%6 - 3 = 6k + 1 + 1 - 3 = 6k - 1.

Thus, if n is part of a twin prime pair and n ≠ 3, it's twin will be n + n%6 - 3.

How it works

Python doesn't have a built-in primality test. While there are short-ish ways to test a single number for primality, doing so for two number would be lengthy. We're going to work with divisors instead.

sum((n+n%6-3)*n%k<1for k in range(2,4*n))

counts how many integers k in the interval [2, 4n) divide (n + n%6 - 3)n evenly, i.e., it counts the number of divisors of (n + n%6 - 3)n in the interval [2, 4n). We claim that this count is 2 if and only if n is part of a twin prime pair.

• If n = 3 (a twin prime), (n + n%6 - 3)n = 3(3 + 3 - 3) = 9 has two divisors (3 and 9) in [2, 12).

• If n > 3 is a twin prime, as seen before, m := n + n%6 - 3 is its twin. In this case, mn has exactly four divisors: 1, m, n, mn.

  Since n > 3, we have m > 4, so 4n < mn and exactly two divisors (m and n) fall into the interval [2, 4n).

• If n = 1, then (n + n%6 - 3)n = 1 + 1 - 3 = -1 has no divisors in [2, 4).

• If n = 2, then (n + n%6 - 3)n = 2(2 + 2 - 3) = 2 has one divisor (itself) in [2, 8).

• If n = 4, then (n + n%6 - 3)n = 4(4 + 4 - 3) = 20 has four divisors (2, 4, 5, and 10) in [2, 16).

• If n > 4 is even, 2, n/2, and n all divide n and, therefore, (n + n%6 - 3)n. We have n/2 > 2 since n > 4, so there are at least three divisors in [2, 4n).

• If n = 9, then (n + n%6 - 3)n = 9(9 + 3 - 3) = 81 has three divisors (3, 9, and 21) in [2, 36).

• If n > 9 is a multiple of 3, then 3, n/3, and n all divide n and, therefore, (n + n%6 - 3)n. We have n/3 > 3 since n > 9, so there are at least three divisors in [2, 4n).

• Finally, if n = 6k ± 1 > 4 is not a twin prime, either n or m := n + n%6 - 3 must be composite and, therefore, admit a proper divisor d > 1.

  Since either n = m + 2 or m = n + 2 and n, m > 4, the integers d, m, and n are distinct divisors of mn. Furthermore, m < n + 3 < 4n since n > 1, so mn has at least three divisors in [2, 4n).