Crittografia-RSA-Fattorizzazione

[P_1_6]

Crittografia-RSA-Fattorizzazione

mi aiuteresti a capire da questo esempio chi sono $d$ e $c$ in questo esercizio di fattorizzazione ?

Quale ordine di grandezza devono avere $d$ e $c$ nei confronti di $p$ ed $N$ e tra di loro?


Se $N=p*q$ e $(p+q-4) mod 8 =0$ e $p >= (p+q)/4$

$N=66390187$

$(3*N-1)/8=24896320$

$((3*N-1)/8-1)/3=8298773$

$8298773-((x+d)-1)*(p-c)=(((p-c)*3*(p-c)*3-1)/8-1)/3-(p-c)*((p-c)-3)/2$
,
$c=1/100000000$
,
$p=[12*x+6-sqrt[3*(48*x^2+48*x+11-8*24896320)]]/3$


variare $d$ in

$d=1/10000000000$
or
$d=2/10000000000$
or
$d=3/10000000000$
or
$d=4/10000000000$
or
$d=5/10000000000$
or
$d=6/10000000000$
or
$d=7/10000000000$
or
$d=8/10000000000$
or
$d=9/10000000000$
or
$d=10/10000000000$

per $d=6/10000000000$ -> $x0=2078$ e $x=2075,.....$


spiegazione aggiuntiva

Each number $H=x*y$ && $(y-x) mod 8 == 0$

$K=(H-1)/8$ e$d n=y-x$

then one of these is true

$1$
$solve (H-1)/8=3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2$
$2$
$solve (H-1)/8=3*x*(x+1)/2-3*y*(y-1)/2+(3*x-1+1)*(3*x-1+2)/2-(x-y+1)$
$3$
$solve (H-1)/8=3*x*(x+1)/2-3*y*(y-1)/2+(3*x-2+1)*(3*x-2+2)/2-2*(x-y+1)$

We observe in particular that in $1$ :

$H =0 mod 3$

therefore $H/3=N=p*q$

when $p=[2*(3*x+1-(x-y+1))+1-(4*y-2)]$ and $q=[2*(3*x+1-(x-y+1))+1]$

and $p+q = 4 mod 8$

$H$ has the characteristic that

$H-1 = 0 mod 8$

and

$((H-1)/8-1)=0 mod 3$

Let's rewrite $p$ as a function of only $x$ where $M=(H-1)/8$

solve
$p=[2*(3*x+1-(x-y+1))+1-(4*y-2)]$
,
$3*x*(x+1)/2-3*y*(y-1)/2+(3*x+1)*(3*x+2)/2=M$

and we get

$p=[12*x+6-sqrt[3*(48*x^2+48*x+11-8*M)]]/3$

$A=((H-1)/8-1)/3$

has the characteristic

$A-(x-1)*p=((p*3*p*3-1)/8-1)/3-p*(p-3)/2$

and in a about $c$ of $p$, $x$ will be very near to $x0$