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let n be an odd integer. Assume that a is an integer such that a=2(modn) show that gcd(a,n) = 1
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Tell me if I'm close. n| a -2 a-2 = nq+0 a = nq+2 --> n=2q+r 0<r<2
hmm, a few examples might help 5 = 2 mod(3) 6 = 2 mod(4); but gcd (4,6) = 2
let n be odd, helps if i read the whole thing
of course, of a and n are even, there are gonna have a gcd of at least 2
2 cases, let a = 2k, and let a = 2k+1, such that n=2m+1 is an idea
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