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Old 2021-07-22, 14:05   #4
RomanM
 
Jun 2021

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Once again, Why does it work? (Personally I don't know yet))

Code:
\p300
{p=233108530344407544527637656910680524145619812480305449042948611968495918245135782867888369318577116418213919268572658314913060672626911354027609793166341626693946596196427744273886601876896313468704059066746903123910748277606548649151920812699309766587514735456594993207;
c=ceil(sqrt(p));
for(n=1,p,
u=c+n; \\initial values
b=lift(Mod(u^2,p));
a=lift(Mod(b^2,p));

for(y=1,250, \\The riddle is here, in this cycle.

t=ceil((b^2-a)^(1/2));
b=lift(Mod(t^2,p));
a=lift(Mod(b^2,p));

if(b<c, break());); \\break at sub sqrt residual
localprec(7);
z=(b/c/1.);
if(z<1,print(z," ",t));
);}
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