Let P(n) be the product of four consecutive integers, starting with n, minus one:
P(n) = n* (n+1)* (n+2) *(n+3) - 1.
Clearly, P(n) can be either prime or not. e.g. P(1) = 23 is prime and P(2) = 119 = 7*17 is not.
Let us call "kc-tuple" a string of k consecutive values : { n, n+1, n+2, ..., n+k-1} such that : P(n), P(n+1), ... , P(n+k-1) are all prime.
There is a significant number of twins (kc = 2) or triplets (kc = 3) but the quadruplets (kc = 4) are rather sparse, and 5 & 6-tuples even sparser.
1) Find the 1st four examples for kc=2,3,4, ( 5 & 6 optional) .
2) Prove there are no kc-tuples for kc>6.
I used Python to find the 1st four occurrences for kc = 2, 3, and 4, and as of this writing, had not found one for kc = 5 or 6 (did I mention Python is slow?). So as of right now, this is what I have:
Part I:
kc = 2: n = 3, 10, 39, and 48
kc = 3: n = 6, 101, 1125, and 1539
kc = 4: n = 7046, 24274, 115811, and 116345
These values are correct assuming my code was written correctly, AND my Miller-Rabin probable prime test isn't allowing composite values to pass through.
Part II:
Let us look at the 4 numbers in P(n), modulo 7. We can have one of seven possible sets:
{0, 1, 2, 3} {1, 2, 3, 4} {2, 3, 4, 5} {3, 4, 5, 6} {4, 5, 6, 0} {5, 6, 0, 1} {6, 0, 1, 2}
Now, the product of any set with a zero, is zero. That is, the product of {n, n+1, n+2, n+3} is a multiple of 7. That means, for these sets, P(n) is congruent to 6 modulo 7. Now, let's look at the three sets without a zero:
{1, 2, 3, 4} {2, 3, 4, 5} {3, 4, 5, 6}
1 * 2 * 3 * 4 = 24; 24 - 1 = 23; 23 = 2 mod 7
2 * 3 * 4 * 5 = 120; 120 - 1 = 119; 119 = 0 mod 7
Since any P(n) where n is 2 modulo 7, will be a multiple of 7, we can see that every 7th set will fail to be prime, so no instances of kc > 6 can exist.
Edited on December 19, 2010, 8:06 pm
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Posted by Justin
on 2010-12-19 16:57:55 |
solution
