Group Effort - Page 2

geoff · 2005-06-22, 01:18

Thanks for that explaination, from thinking about it I have worked out this proof that 159986*5ⁿ+1 and 346802*5ⁿ-1 are never prime:

For integers k and c we can write k*5ⁿ+c = k*5^12m+r+c = k*5^r(5^12m-1) + (k*5^r+c) = A(k,c) + B(k,c), where n=12m+r, r < 12.

Since x^am-1 = (x^a-1)(x^a(m-1) + x^a(m-2) + ... + 1) when m > 0, 5¹²-1 divides 5^12m-1 for any m >= 0, and thus divides A(k,c).

By direct calculation {3,7,13,31,601} all properly divide 5¹²-1 and hence properly divide A(k,c), and (there are 24 cases) one of {3,7,13,31,601} always properly divides B(159986,1) and B(346802,-1) as well.

Therefore 159986*5ⁿ+1 = A(159986,1) + B(159986,1) and 346802*5ⁿ-1 = A(346802,-1) + B(346802,-1) are always properly divisible by one of {3,7,13,31,601} and hence not prime.

robert44444uk · 2005-06-22, 12:12

Geoff, thanks for that.

How do we show that there are 24 permutations of the five primes? And why is it that the particular permutation gives the lowest k?

Can you also put in mathematical notation why 12 (and not 10 or someo ther number, say 4 or 6) is the lowest composite which satisfies the factor requirement and the sigma requirement?

I think we need this last thing to extend the proof to demonstrate the next stage, that any other covering set has to cover greater than 12 values of n.

Regards

Robert

geoff · 2005-06-23, 05:01

Quote:

How do we show that there are 24 permutations of the five primes? And why is it that the particular permutation gives the lowest k?

The 24 cases mentioned are the particular values of B(k,c) = k*5^r+c that you get when you set k=159986,c=1 and k=346802,c=-1 and let r range from 0 to 11 in either case. The calculation is just checking that each of these 24 numbers has a factor in the set {3,7,13,31,601}. I will try to find a clearer way to express this.

Quote:

Can you also put in mathematical notation why 12 (and not 10 or someo ther number, say 4 or 6) is the lowest composite which satisfies the factor requirement and the sigma requirement?

By the Division Algorithm, for any of a > 0 (a=12 in our case) we can set n=am+r where r < a, and write kbⁿ+c as the sum kb^r(b^am-1) + (kb^r+c).

For this proof to work, we need a sequence of a terms kb^r+c, 0 <= r < a, where each term has a common factor with b^a-1, (because b^a-1 is itself a factor of b^am-1).

Showing that no such sequence exists for b = 5 when c = 1 and k < 159986 or when c = -1 and k < 346802 is a finite computation, but would only show that this particular proof won't work for any smaller value of a.

wblipp · 2005-06-23, 13:51

Quote:

Originally Posted by geoff

I will try to find a clearer way to express this.

Have you see the Team Prime Rib proof that 78557 is a Sierpinski number (base 2)? I suggest using that as a template. It has a main page that shows the partitions and a link for each partition with the details.

http://sob.arsfoodcourt.com/index.ph...isplay&ceid=44

geoff · 2005-06-25, 14:07

Here is a more explicit version of the proof that 159986*5^n+1 is never prime. The proof for 346802*5^n-1 is the same except for the particular numbers.

The proof uses two simple facts: if A and B are both properly divisible by C then A+B is also properly divisible by C; and for any integer n >=0 and a > 0 we can write n = am + r for some integers m >=0 and r < a.

Note first that:

159986*5^0+1 = (3) * 17 * 3171
159986*5^1+1 = 11 * 11 * 11 * (601)
159986*5^2+1 = (3) * 29 * 31 * 1483
159986*5^3+1 = (7 * 13 * 219761
159986*5^4+1 = (3) * 3 * 41 * 233 * 1163
159986*5^5+1 = (31) * 16127621
159986*5^6+1 = (3) * 11 * 1153 * 65699
159986*5^7+1 = (13) * 961454327
159986*5^8+1 = (3) * 31 * 671984207
159986*5^9+1 = (7) * 193 * 811 * 285191
159986*5^10+1 = (3) * 3 * 3 * 3 * 107 * 180265753
159986*5^11+1 = 11 * (13) * 31 * 1762196347

Also note that 5^12-1 = 2 * 2 * 2 * 2 * 3 * (3) * (7) * (13) * (31) * (601)

For any n >= 0 set n = 12m + r where m >=0 and r < 12. Then substituting 12m+r for n in 159986*5^n+1 and expanding gives

159986*5^(12m+r)+1 = [159986*5^r*(5^12-1)*(5^(m-1)+5^(m-2)+...+1)] + [159986*5^r+1]

Each term in square brackets on the right of this sum must be properly divisible by one of the primes 3,7,13,31 or 601 because (5^12-1) is properly divisible by all of them and (159986*5^r+1) is divisible by at least one of them, as noted above.

Therefore the sum on the left, 159986*5^n+1, is also properly divisible by at least one of the primes 3,7,13,31 or 601 and so is not prime.

2005-06-22, 01:18	#12
geoff Mar 2003 New Zealand 13×89 Posts	Thanks for that explaination, from thinking about it I have worked out this proof that 1599865ⁿ+1 and 3468025ⁿ-1 are never prime: For integers k and c we can write k5ⁿ+c = k5^12m+r+c = k5^r(5^12m-1) + (k5^r+c) = A(k,c) + B(k,c), where n=12m+r, r < 12. Since x^am-1 = (x^a-1)(x^a(m-1) + x^a(m-2) + ... + 1) when m > 0, 5¹²-1 divides 5^12m-1 for any m >= 0, and thus divides A(k,c). By direct calculation {3,7,13,31,601} all properly divide 5¹²-1 and hence properly divide A(k,c), and (there are 24 cases) one of {3,7,13,31,601} always properly divides B(159986,1) and B(346802,-1) as well. Therefore 1599865ⁿ+1 = A(159986,1) + B(159986,1) and 3468025ⁿ-1 = A(346802,-1) + B(346802,-1) are always properly divisible by one of {3,7,13,31,601} and hence not prime. Last fiddled with by geoff on 2005-06-22 at 01:50 Reason: n=12m+r, not m+r

2005-06-22, 12:12	#13
robert44444uk Jun 2003 Oxford, UK 7×277 Posts	Cases Geoff, thanks for that. How do we show that there are 24 permutations of the five primes? And why is it that the particular permutation gives the lowest k? Can you also put in mathematical notation why 12 (and not 10 or someo ther number, say 4 or 6) is the lowest composite which satisfies the factor requirement and the sigma requirement? I think we need this last thing to extend the proof to demonstrate the next stage, that any other covering set has to cover greater than 12 values of n. Regards Robert

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2005-06-25, 14:07	#16
geoff Mar 2003 New Zealand 2205₈ Posts	Here is a more explicit version of the proof that 1599865^n+1 is never prime. The proof for 3468025^n-1 is the same except for the particular numbers. The proof uses two simple facts: if A and B are both properly divisible by C then A+B is also properly divisible by C; and for any integer n >=0 and a > 0 we can write n = am + r for some integers m >=0 and r < a. Note first that: 1599865^0+1 = (3) 17 * 3171 1599865^1+1 = 11 11 * 11 * (601) 1599865^2+1 = (3) 29 * 31 * 1483 1599865^3+1 = (7 13 * 219761 1599865^4+1 = (3) 3 * 41 * 233 * 1163 1599865^5+1 = (31) 16127621 1599865^6+1 = (3) 11 * 1153 * 65699 1599865^7+1 = (13) 961454327 1599865^8+1 = (3) 31 * 671984207 1599865^9+1 = (7) 193 * 811 * 285191 1599865^10+1 = (3) 3 * 3 * 3 * 107 * 180265753 1599865^11+1 = 11 (13) * 31 * 1762196347 Also note that 5^12-1 = 2 * 2 * 2 * 2 * 3 * (3) * (7) * (13) * (31) * (601) For any n >= 0 set n = 12m + r where m >=0 and r < 12. Then substituting 12m+r for n in 1599865^n+1 and expanding gives 1599865^(12m+r)+1 = [1599865^r(5^12-1)(5^(m-1)+5^(m-2)+...+1)] + [1599865^r+1] Each term in square brackets on the right of this sum must be properly divisible by one of the primes 3,7,13,31 or 601 because (5^12-1) is properly divisible by all of them and (1599865^r+1) is divisible by at least one of them, as noted above. Therefore the sum on the left, 1599865^n+1, is also properly divisible by at least one of the primes 3,7,13,31 or 601 and so is not prime.