## Factorization of 6 consecutive 2139-digit integers

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• Application of a Prouhet-Tarry-Escott solution: factorization of 6 consecutive 2139-digit integers Let b = 5026578700; m = 13416739015680*b; x =
Message 1 of 4 , Dec 19, 2009
Application of a Prouhet-Tarry-Escott solution:
factorization of 6 consecutive 2139-digit integers

Let b = 5026578700; m = 13416739015680*b;
x = (2^67+28683395)*b; p25 = 6805713408228112745527499;
y = (8*x^6+24*x^5+50*x^4+54*x^3+41*x^2+12*x-148)^2;
N = (y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-3;
then
p2088 = N/(3*5*115523*2364592753*16987001959*p25);
p2131 = (N+1)/(2^2*7^2*462899);
p2132 = (N+4)/(13*652739);
p2116 = (N+5)/(2^5*5^2*37*241*28843*794200385809);
are prime and the factorizations of (N+2) and (N+3) in
contain no prime with more than 172 digits.

Comments: This sets a record at http://tinyurl.com/ycxabks
for both 5 and 6 consecutive factorizations. The algebraic
factorization of (N+2)*(N+3) comes from a sextic
substitution in Prouhet-Tarry-Escott solution 10.12.151 of
http://www.math.uni.wroc.pl/~jwr/PTE resulting in 4 linear
factors, 5 quadratics, 12 cubics, a quartic and 15 sextics.
OpenPFGW and GMP-ECM found 44 triplets of probable primes,
each promising a run of 5 factorizations after NFS effort on
sextics. One was extended to a run of 6, by extracting the
prime p25. After using ECM and Msieve's MPQS, 9 composite
factors of (N+2)*(N+3) remained for GGNFS. Two composites
had GNFS difficulty less than 122, leaving 7 sextics with
SNFS difficulty 180. Pari-GP and Primo proved primality of
factors below 180 digits and above 2000 digits,
respectively. Neither of the composite numbers
c2117 = (N - 1)/(2*6911*18959*675413*81134281)
c2118 = (N + 6)/(3*167858252466989944843)
is likely to have a prime divisor with less than 30 digits.

• ... Congratulations to both! http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm is now updated. -- Jens Kruse Andersen
Message 2 of 4 , Dec 21, 2009
> Application of a Prouhet-Tarry-Escott solution:
> factorization of 6 consecutive 2139-digit integers
>

Tom wrote:
> Factorization of 4 consecutive 5257-digit numbers
> Let n = 297079965*2^17434-1
>
> n = 4363*22567639*p5246
> n+1 = 297079965*2^17434 = 2^17434*3^2*5*7*13*72547
> n+2 = 297079965*2^17434+1 (prime)
> n+3 = 297079965*2^17434+2 = 2*(297079965*2^17433+1) (2*prime)

Congratulations to both!
http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm
is now updated.

--
Jens Kruse Andersen
• Application of a multigrade equality found by Tarry: factorization of 8 consecutive 850-digit integers Let x = 429*(2^50 + 9096237); y = (16*x^6 - 72*x^4 +
Message 3 of 4 , Dec 22, 2009
Application of a multigrade equality found by Tarry:
factorization of 8 consecutive 850-digit integers

Let x = 429*(2^50 + 9096237);
y = (16*x^6 - 72*x^4 + 81*x^2 - 25)^2;
N = 72*(y - 2^2)*(y - 2^8)*(y - 5^4)*(y - 21^2)/14! - 3;
then
p849 = N/3;
p812 = (N + 1)/(2*9277*9862129*3520556383*110169881898372133);
p850a = (N + 2);
p850b = (N + 5)/2;
p845 = (N + 6)/(3*19*1009);
p844 = (N + 7)/(2^2*208577);
are prime and the factorizations of (N + 3) and (N + 4) in
contain no prime with more than 108 digits.

Comments: Algebraic factorization of (N + 3)*(N + 4) comes from
a sextic substitution in a multigrade equality discovered in
1913 by Gaston Tarry, amateur extraordinaire:
http://www-history.mcs.st-and.ac.uk/Biographies/Tarry.html
OpenPFGW, GMP-ECM, Pari-GP and Primo achieved the rest.
Remarkably, p849, p850a and p850b have BLS proofs, with
helpers of less than 16 digits from the factorization of
N + 3 = 3*(p849 + 1) = p850a + 1 = 2*(p850b - 1).
Neither of the composites
c837 = (N - 1)/(2^2*95819*27609121);
c824 = (N + 8)/(5*311137*114685080442874011127);
is likely to have a prime divisor with less than 25 digits.