Why 37 Comes Home

The polynomial f(p) = p^2 - p - 1 maps primes to integers. Call f(p) '11-smooth' if all its prime factors come from {2, 3, 5, 7, 11}. Among 1,229 primes below 10,000, exactly three produce smooth values: f(2) = 1, f(3) = 5, and f(37) = 1331 = 11^3. Two are from the ring's own prime set. The third -- 37 -- is the only outsider whose output lands back inside.

What Happens at Each Prime

Apply f(p) = p^2 - p - 1 to the ring's primes and then to nearby primes. A value is '11-smooth' if all its factors come from {2, 3, 5, 7, 11}:

The pattern breaks immediately after 2 and 3. Primes 5, 7, 11, and 13 all produce values with factors outside {2,3,5,7,11}. Then at p = 37 -- the 12th prime, well past the ring's own primes -- the output is 11^3, purely smooth. No other prime below 10,000 does this.

Why 37?

37 is the 12th prime. 12 = 2^2 * 3 is the Carmichael function lambda(Z/210) -- the period of the smallest ring built from {2, 3, 5, 7}. Its Cunningham half (37-1)/2 = 18 = 1+2+3+5+7, the sum of the five chain elements up to 7.

37 also decomposes as (2*3)^2 + 1 = 36 + 1, or equivalently 2^5 + 5 = 32 + 5. Both expressions use only primes from {2, 3, 5}.

In Z/12,612,600, the number 37 has multiplicative order 420 -- the full Carmichael period. This makes it a primitive element: every unit in the ring is a power of 37. It is coprime to every prime dividing the ring, so it appears in all six CRT channels simultaneously.

Where 1331 Lands

The output f(37) = 1331 = 11^3 has CRT decomposition (3, 8, 6, 8, 0, 5) in Z/12,612,600. The mod-11 channel is zero: 11^3 is divisible by 11, so it annihilates its own channel. All other channels are nonzero. The polynomial maps a primitive element to a number that lives entirely in one prime's domain.

Explore: f(p) = p^2 - p - 1

Enter any prime p to compute f(p) = p^2 - p - 1 and check if it is 11-smooth.

Enter prime p:

Try: p=2 (gives 1), p=3 (gives 5), p=37 (gives 11^3 = 1331 -- the only outsider!), p=5 (19, not smooth), p=7 (41, not smooth).

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