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Old 2021-08-01, 13:48   #16
Alberico Lepore
 
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May 2017
ITALY

2·32·29 Posts
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Code:
check=0

i=0

while(i<10) {

solve this system

sqrt(N/((10+i)/10))=a 
, 
((10+i)/10*a+a-4)/8=x 
, 
2*x*(x+1)-b*(b-1)/2=(N-3)/8 

}

memorize b(i)


C=0

while (check==0){

i=0

while(i<10 && check==0) {

h=x-b(i)/2-C  // b/2 must be integer
,
[2*(h-1)*(h-1+1)]  
< 
N-3)/8-b(i)/2*(4*x+1-2*(y-1)) 
<=  
[2*h*(h+1)] 
,  
2*(x)*(x+1)-y*(y-1)/2=(N-3)/8


while(min_h <= max_h && check==0){

x=h+b/2+C

2*(x)*(x+1)-y*(y-1)/2=(N-3)/8

calculate p

p=4*x+1-2*(y-1)

if(N mod p ==0 && p!=1 && p!=N){
check=1
break

}

min_h++

}
i++
}

C++
}



Example

N=390644893234047643
,
sqrt(N/(15/10))=a
,
(15/10*a+a-4)/8=x
,
2*x*(x+1)-b*(b-1)/2=(N-3)/8

b = 63790420






h=x-(63790420)/2-C
,
[2*(h-1)*(h-1+1)]
<
(390644893234047643-3)/8-(63790420)/2*(4*x+1-2*(y-1))
<=
[2*h*(h+1)]
,
2*(x)*(x+1)-y*(y-1)/2=(390644893234047643-3)/8


per C=-7454

127855236<=h<127855255 -> size range = 19


per h=127855241
,
h=x-(63790420)/2-7454

->

x=159757905



the problem is that size range of h is decreasing

for C=0

127580838<=h<127584034 -> size range = 3196

for C=3727

127775161<=h<127775188 -> size range =27


an exponential decrease would seem to our advantage




*********************************************************************

UPDATE:

I tried to solve in x and I noticed that the first valid value is our x = 159757905
if
it would always happen

the cost of factoring 390644893234047643 would be 7454 * 10 = 74540

Tomorrow morning I will continue with other tests

https://www.wolframalpha.com/input/?...%2F8+%2Ch+%2Cy

Last fiddled with by Alberico Lepore on 2021-08-01 at 19:00 Reason: UPDATE
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