Let x and y be integers. Prove that if there is an equivalence class [a] modulo

Question

Let x and y be integers. Prove that if there is an equivalence class [a] modulo n such that x is an element of [a] and y is an element of [a], then (x,n)=(y,n).

Explanation / Answer

Let z=r(mod6), then z-r=6m=>...( where 6m is the multiple of 6) z=6m+r=> r is the remainder when z is divided by 6=> r=0,1,2,3,4,5. Let i be any of 0,1,2,3,4,5, then if z-(i+6)=6m, then z-i=6m+6=> z-i=6(m+1)=> z=i (mod 6)=> r must not be >5. z=i (mod 6) has a general solution for z such that z(i)={ t | t=i+6k, t,k are integers }, for i+6k-i=6k is a multiple of 6. Suppose that z(i) & z(j), where 0= i-j=6(k'-k) since i, j < 6 it is impossible that i-j is a multiple (+/-) of 6. So, z(i) =/= z(j) & there are exactly 6 different classes of z(i), i=0,1,2,3,4,5.

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