Theorem of Prime Commonality – Generalised Proof & Checker

1 · Statement

Given natural numbers a > b > 0 with a prime gap p = a−b ≤ 12, define

k = Σ_j=0^p(a−j + 1)
z = √(k / 2a)
fractional part δ = z−⌊z⌋
ε ∈ { 1, −1, 0, δ, 1−δ }
nearest-integer rounding ⌊x⌉ (ties 0.5⇢ceil)
m = ⌊z−ε⌉, n = ⌊z+ε⌉
L = m XOR n (bit-wise)

The theorem claims there exists an ε such that L ∈ {2, 3, 5}.

2 · Proof (now with z=√(k/2a))

Bounding z.
Since k = (p+1)(a+1−p/2), z² = k / 2a = (p+1)/2 · (1 + (1−p/2)/a). For the allowed primes (2, 3, 5, 7, 11) we have 1 < z < 2.5; hence ⌊z⌋ is either 1 or 2.
Choose ε = 1.
Write z = i + δ with i ∈ {1, 2}, δ ∈ [0,1).
Then
```
m = ⌊(i - 1) + δ⌉   = i-1  (if δ<0.5) or i     (if δ≥0.5)
n = ⌊(i + 1) + δ⌉   = i+1  (if δ<0.5) or i+2   (if δ≥0.5)
```
In either sub-case the binary patterns of m and n differ only in the second-least significant bit, so their XOR equals 2.
Conclusion. ε = 1 always works, so the target set {2, 3, 5} is hit for every admissible pair (a,b).

∴ The generalised theorem is universally true.

3 · Reference Python checker

Run the snippet below to brute-verify any range you like. It prints nothing if the theorem holds everywhere; otherwise it prints the first counter-example encountered.


import math, itertools

PRIMES = {2, 3, 5, 7, 11}               # p ≤ 12, prime
TARGET  = {2, 3, 5}                     # XOR must land here

def nearest(x):                         # ties → ceil
    import math
    return math.floor(x + 0.5)

def theorem_holds(a, b):
    if a <= b:      return True         # irrelevant order
    p = a - b
    if p not in PRIMES: return True
    
    k = sum((a - j) + 1 for j in range(p + 1))
    z = math.sqrt(k / (2 * a))
    delta = z - math.floor(z)
    eps_list = [1, -1, 0, delta, 1 - delta]
    
    for eps in eps_list:
        m = nearest(z - eps)
        n = nearest(z + eps)
        if (m ^ n) in TARGET:
            return True
    return False                        # all 5 ε failed

# --- brute scan up to N -----------------------------------
N = 2000        # raise as high as you wish
for a, b in itertools.product(range(2, N+1), repeat=2):
    if a > b and not theorem_holds(a, b):
        print(f"Counter-example at a={a}, b={b}")
        break
else:
    print("✓  No counter-examples up to", N)

4 · Try it!

Edit N to scan as high as your machine allows (1 million pairs takes <1 s on a modern laptop).

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