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d1= pgcd(56,35)
x y 56x+ 35y=d1
d2= pgcd(309,186), d3= pgcd(1024,729), d4= pgcd(12075,4655).
a b d pgcd
u v
au +bv =d.
u v
(?)d= pgcd(a, b) (u0, v0)∈Z2
au0+bv0=d(u, v)au +bv =d(uk, vk)
k∈Z∗
(uk=u0+kb1,
vk=v0−ka1
a1b1a=da1b=db1
Z2
5x−18y= 4,6x+ 15y= 28.
a, b, c d
(pgcd(a, b) = d)⇐⇒ (pgcd(ac, bc) = dc)
(pgcd(a, b)=1
pgcd(a, c)=1 =⇒(pgcd(a, bc) = 1)
(pgcd(a, b) = 1) ⇐⇒ ∀m≥2,∀n≥2,(pgcd(am, bn) = 1)
(pgcd(a, b) = 1) ⇐⇒ (pgcd(a+b, ab) = 1)
(pgcd(a, b) = d)⇐⇒ ∀n≥2,(pgcd(an, bn) = dn)
(m, n)∈Z24n4+ 1 = 3m2+ 3