Conjecture. If $n \in \Bbb{N}\setminus 1$ is not a twin prime average, then $n^2 - 1$ is not square-free.

Question

From the evidence:

from sympy import *

N = 1000

for n in range(2, N):
   if not isprime(n + 1) or not isprime(n-1):
      s = n**2 - 1
      if rad(s) != s:
         print(f"True for {n} = n, n^2 - 1 = {s}")
      else:
         print(f"False for {n} = n, n^2 - 1 = {s}")

Prints all true:

.........

True for 963 = n, n^2 - 1 = 927368
True for 964 = n, n^2 - 1 = 929295
True for 965 = n, n^2 - 1 = 931224
True for 966 = n, n^2 - 1 = 933155
True for 967 = n, n^2 - 1 = 935088
True for 968 = n, n^2 - 1 = 937023
True for 969 = n, n^2 - 1 = 938960
True for 970 = n, n^2 - 1 = 940899
True for 971 = n, n^2 - 1 = 942840
True for 972 = n, n^2 - 1 = 944783
True for 973 = n, n^2 - 1 = 946728
True for 974 = n, n^2 - 1 = 948675
True for 975 = n, n^2 - 1 = 950624
True for 976 = n, n^2 - 1 = 952575
True for 977 = n, n^2 - 1 = 954528
True for 978 = n, n^2 - 1 = 956483
True for 979 = n, n^2 - 1 = 958440
True for 980 = n, n^2 - 1 = 960399
True for 981 = n, n^2 - 1 = 962360
True for 982 = n, n^2 - 1 = 964323
True for 983 = n, n^2 - 1 = 966288
True for 984 = n, n^2 - 1 = 968255
True for 985 = n, n^2 - 1 = 970224
True for 986 = n, n^2 - 1 = 972195
True for 987 = n, n^2 - 1 = 974168
True for 988 = n, n^2 - 1 = 976143
True for 989 = n, n^2 - 1 = 978120
True for 990 = n, n^2 - 1 = 980099
True for 991 = n, n^2 - 1 = 982080
True for 992 = n, n^2 - 1 = 984063
True for 993 = n, n^2 - 1 = 986048
True for 994 = n, n^2 - 1 = 988035
True for 995 = n, n^2 - 1 = 990024
True for 996 = n, n^2 - 1 = 992015
True for 997 = n, n^2 - 1 = 994008
True for 998 = n, n^2 - 1 = 996003
True for 999 = n, n^2 - 1 = 998000

It seems that every non-twin prime average $n\gt 1$ is such that $n^2 - 1$ (the product of its would-be primes) is not square-free.

How can we prove this by elementary means?

---

from sympy import *

N = 100000
k = 2

for n in range(3, N):
   if not isprime(n + k) or not isprime(n - k):
      s = n**2 - k**2
      if rad(s) != s:
         print(f"True for {n} = n, n^2 - k^2 = {s}")
      else:
         print(f"False for {n} = n, n^2 - k^2 = {s}")
         assert(0)

According to the last code output, I conclude with this conjecture:

Generalized Conjecture. For all fixed $k \geq 1$, for sufficiently large $n \in \Bbb{N},$ namely about $n - k \gt 0$, we have that $n \pm k \in \Bbb{P}$ or in other words $n$ is a $2k$-separated prime average if and only if $n^2 - k^2$ is a square-free integer!

How can we prove it or has this already been proved?

Answer 1

Your conjecture is not true. $n=14$ is not a twin prime average (as $14+1 = 15 = 3\cdot 5$ is not a prime), but $n^2 - 1 = 195 = 3\cdot 5\cdot 13$ is square-free.

You could sharpen your conjecture by requiring that neither $n-1$ nor $n+1$ are prime. But it still won't be true. Now $n=34$ with $n-1 = 3\cdot 11$ composite, $n+1 = 5\cdot 7$ composite and $n^2 - 1 = 1155 = 3\cdot 5\cdot 7\cdot 11$ squarefree is a counterexample.