∀a,b∈Z,∃x,y∈Z,ax+by=gcd(a,b)\forall a,b \in\mathbb{Z},\exist x,y\in\mathbb{Z},ax+by=\gcd(a,b)∀a,b∈Z,∃x,y∈Z,ax+by=gcd(a,b)
设 g=gcd(a,b)g = \gcd(a,b)g=gcd(a,b),则有 k1g=a,k2g=b,gcd(k1,k2)=1k_1g =a,k_2g=b,\gcd(k_1,k_2)=1k1g=a,k2g=b,gcd(k1,k2)=1
k1gx+k2gy=gk_1gx+k_2gy=gk1gx+k2gy=g
k1x+k2y=1k_1x+k_2y=1k1x+k2y=1
k1x−1=−yk2k_1x-1=-yk_2k1x−1=−yk2
k1x≡1(modk2)k_1x\equiv1 \pmod {k_2}k1x≡1(modk2)
根据欧拉定理,取 x=k1φ(k2)−1x=k_1^{\varphi(k_2)-1}x=k1φ(k2)−1,则 yyy 显然存在。