## Factorization of 8 consecutive 600-digit integers

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• Let x=(1320*(10^22+1932187))^3 y=(x*(5*x+9)/2-31)^2 N=(y-23^2)*(y-24^2)/55440-6 then N+k is factorized for k in [0,7]. Proof:
Message 1 of 4 , Sep 2, 2007
Let

x=(1320*(10^22+1932187))^3
y=(x*(5*x+9)/2-31)^2
N=(y-23^2)*(y-24^2)/55440-6

then N+k is factorized for k in [0,7].

Proof:

Software used:
GGNFS (with SNFS difficulty = 151.4)
GMP-ECM
Msieve
OpenPFGW
Pari-GP
Primo

Neighbouring composites:
c569=(N-1)/(2^2*25711801*125018686680668134578227)
c597=(N+8)/(5*281)
subjected to 400 ECM curves at B1=250000

2 September 2007
• ... Congratulations! http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm is updated once again. ? x=u^3; ? y=(x*(5*x+9)/2-31)^2; ?
Message 2 of 4 , Sep 2, 2007
> x=(1320*(10^22+1932187))^3
> y=(x*(5*x+9)/2-31)^2
> N=(y-23^2)*(y-24^2)/55440-6
>
> then N+k is factorized for k in [0,7].

Congratulations!
http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm
is updated once again.

? x=u^3;
? y=(x*(5*x+9)/2-31)^2;
? N=(y-23^2)*(y-24^2)/55440-6;
?
? factor(N+2)
%108 =
[u^3 + 4 1]
[u^3 + 5 1]
[u^3 + 6 1]
[5*u^3 - 21 1]
[5*u^3 - 16 1]
[5*u^3 - 11 1]
[5*u^6 + 9*u^3 + 2 1]

? factor(N+3)
%109 =
[u 3]
[u^3 - 2 1]
[5*u^3 + 9 1]
[5*u^3 + 19 1]
[5*u^6 + 9*u^3 - 124 1]
[5*u^6 + 9*u^3 - 86 1]

? factor(N+6)
%110 =
[u - 1 1]
[u^2 + u + 1 1]
[5*u^3 + 14 1]
[5*u^6 + 9*u^3 - 110 1]
[5*u^6 + 9*u^3 - 108 1]
[5*u^6 + 9*u^3 - 16 1]

Nice.

--
Jens Kruse Andersen
• ... It is possible to do slightly better and reduce the number of irreducible sextics to 5/12, instead of the 6/12 that I used. The answer is (Euler s?) 41: ?
Message 3 of 4 , Sep 4, 2007
--- In primeform@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@...> wrote:

> ? x=u^3;
> ? y=(x*(5*x+9)/2-31)^2;
> ? N=(y-23^2)*(y-24^2)/55440-6;
...
> [5*u^6 + 9*u^3 - 16 1]
>
> Nice.

It is possible to do slightly better and reduce
the number of irreducible sextics to 5/12,
instead of the 6/12 that I used.

? y=(x^3/41)^2
? N=(y-41)*(y-41^2)*(y-31^2)*(y-51^2)/141926400000;
? factor(N*(N+1)*(N+4))
%3 =
[x^2 - 41 1]

[x^3 - 2091 1]

[x^3 - 1681 1]

[x^3 - 1271 1]

[x^3 - 1189 1]

[x^3 - 779 1]

[x^3 - 369 1]

[x^3 + 369 1]

[x^3 + 779 1]

[x^3 + 1189 1]

[x^3 + 1271 1]

[x^3 + 1681 1]

[x^3 + 2091 1]

[x^4 + 41*x^2 + 1681 1]

[x^6 - 4305041 1]

[x^6 - 3834361 1]

[x^6 - 3767121 1]

[x^6 - 3027481 1]

[x^6 - 674081 1]

but then the coefficients are larger in SNFS.

David
• The Prouhet-Tarry-Escott chain for the records with k=8,9,10 is very simple. We simply use products of the 3 complex Gaussian primes with smallest odd norm: ?
Message 4 of 4 , Sep 8, 2007
The Prouhet-Tarry-Escott chain for the records with k=8,9,10
is very simple. We simply use products of the 3 complex
Gaussian primes with smallest odd norm:

? a=(2+I)*(3+2*I)*(4+I)
%1 = 9 + 32*I
? b=(2+I)*(3+2*I)*(4-I)
%2 = 23 + 24*I
? c=(2+I)*(3-2*I)*(4-I)
%3 = 31 - 12*I
? 4*imag(c^2)^2==imag(b^2)^2+3*imag(a^2)^2
%4 = 1

Unfortunately, d=(2+I)*(3-2*I)*(4+I)=33+4*I is of no use, here.

David (trying to relate everything to Hardy and Wright)
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