Can I ask for help with a proof?

[kjartan]

I'm trying to prove something that looks like it should be easy, but I keep getting frustrated . . .

Namely:

If a is congruent to b modulo n, then gcd(a,n) = gcd(b,n)




Thanks for any replies!

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[6 cents]

I'm trying to prove something that looks like it should be easy, but I keep getting frustrated . . .

Namely:

If a is congruent to b modulo n, then gcd(a,n) = gcd(b,n)




Thanks for any replies!

assume gcd (a,n) = r
assume gcd (b,n) = s

b = a + t*n (definition of congruent)

Divide both sides by r

b/r = a/r + t*n/r

Since r divides n and a, r must divide b or you would have a non integer equal the sum of two integers. Since r divides both b and n, r must divide s or s would not be the gcd(b,n).

Likewise write a = b - t*n and divide each side by s to show that s must divide r by the same reasoning.

numbers that divide each other are equal.

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