## Factorization of 8 consecutive 850-digit integers

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• 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 1 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.