Start with some small examples, substituting (6k-1) and (6k+1) for p:
Factoring for k=2,3,.. with the number in brackets representing the resulting p:
2=2^5×3×5×7×17×31 (11) 2^5,3,5,7,17,31
3=2^7×3^2×7×41×73 (17) 2^5,3,7 eliminating non-repeats
4=2^6×3×7×11×17×19×31 (23) 2^5,3,7
5=2^5×3×5×7×211×839 (29) 2^5,3,7
6=2^5×3^2×17×307×1223 (35) 2^5,3 eliminating non-repeats
7=2^6×3×5×7×23×73×421 (41) 2^5,3
8=2^7×3×7×23×79×2207 (47) 2^5,3
(6k+1)^6-7(6k+1)^2+6
Factoring for k=2,3,..
2=2^5×3×7×43×167 (13) 2^5,3,7,43,167
3=2^5×3^2×5×7×13×359 (19) 2^5,3,7 eliminating non-repeats
4=2^6×3×7×13×89×157 (25) 2^5,3,7
5=2^8×3×5×7×137×241 (31) 2^5,3,7
6=2^5×3^2×7^3×19×1367 (37) 2^5,3,7
7=2^5×3×7×11×463×1847 (43) 2^5,3,7
8=2^7×3×5^2×601×2399 (49) 2^5,3 eliminating non-repeats
So empirically it seems the greatest common divisor is 2^5*3 or 96. We don't really care too much about primality for this purpose as the factors are clearly cyclic, except for the power of 2.
We can then, with difficulty, factor the expressions.
(6k-1)^6-7(6k-1)^2+6 = 48k(3k-1)(324k^4-216k^3+63k^2-9k-1), so we get 48 at once, and the 5th power of 2 either from k itself being even, or (3k-1) being even for k odd.
Similarly,
(6k+1)^6-7(6k+1)^2+6= 48k(3k+1(324k^4+216k^3+63k^2+9k-1) to which the same observation applies.
However, there is one additional twist. The 6th entry of the (6k-1) form and the 8th entry of the (6k+1) form do not have prime p: they are 35 and 49.
On closer inspection, (6(7k-1)-1) = 7(6n-1) and (6(7k+1)+1) = 7(6n+1), so such entries will never be prime. Those entries can therefore be ignored, with the result that 7 is always a factor of the GCD for prime p, though not all p.
For completeness:
(324k^4-216k^3+63k^2-9k-1) is divisible by 7 for k={1,2,3,4} mod7 (easily shown by substitution). It is NOT so divisible for k={5,6,0} mod7, but if k=0,mod7 then the 48k term is so divisible, and if k=5, mod7, then (3(7k+5)-1) = 7(3k+2), always so divisible, leaving k=5, mod7, which however can never be prime as already shown.
In like manner,
(324k^4+216k^3+63k^2+9k-1) is divisible by 7 for k={3,4,5,6} mod7 (easily shown by substitution). It is NOT so divisible for k={1,2,0} mod7, but if k=0,mod7 then the 48k term is so divisible, and if k=2, mod7, then (3(7k+2)+1) = 7(3k+1), always so divisible, leaving k=1, mod7, which however can never be prime as already shown.
So the final result is that 2^5*3*7 or 672 is the GCD of all such prime numbers.
This turned out to be a good 'adventure'!
Edited on May 4, 2024, 5:47 am
|
|
Posted by broll
on 2024-05-03 10:43:10 |
Possible Solution
