if a≡b mod n prove that gcd(a,n)=gcd (b,n)
please setps by steps because i just learning it.
∵ a≡b(mod n)
a=b+nt Þ (a,b)=(b,n)
∵ b≡a(mod n) b=a+ns Þ (b,a)=(a,n)
又 (a,b)=(b,a)∴ (a,n)=(a,b)=(b,n)得證
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∵ a≡b(mod n)
a=b+nt Þ (a,b)=(b,n)
∵ b≡a(mod n) b=a+ns Þ (b,a)=(a,n)
又 (a,b)=(b,a)∴ (a,n)=(a,b)=(b,n)得證