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Old 2020-11-19, 22:44   #17
Kebbaj
 
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"Kebbaj Reda"
May 2018
Casablanca, Morocco

22×19 Posts
Default for a = 3 the numbers found all end with 63

Quote:
Originally Posted by paulunderwood View Post
This script outputs no counterexamples:

Code:
{
forstep(a=3,100,1,print(a);
D=a^2-4;E=D^2-4;
forstep(n=4*E-1,50000000*E,4*E,
if(kronecker(D,n)==-1&&!ispseudoprime(n),
r=Mod(D,n)^((n-1)/2);s=Mod(E,n)^((n-1)/2);
if(r==-1&&s==1,
print([n,r,s])))))
}
Can you find one?
for a = 3 the numbers found all end with 63:
[177655186463, 3]
[286748218763, 3]
[340473667463, 3]
so I added an "if ((n-floor (n / 100) * 100) == 63" , to find only the numbers that end with 63.

Code:
{
forstep(a=3,3,1,print(a);
D=a^2-4;E=D^2-4;
forstep(n=177655186463,1840473667452,4*E,
if((n-floor(n/100)*100)==63,
if(kronecker(D,n)==-1&&!ispseudoprime(n),
r=Mod(D,n)^((n-1)/2);s=Mod(E,n)^((n-1)/2);
if(r==-1&&s==1,print([n,r,s]))))))
}
There are no other numbers up to 1840473667452.
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