Proof: concerning congruent modulo.

[RichardParker]

Homework Statement

Suppose a, b, c, d ∈ Z and n ∈ N. If a ≡ b (mod n) and c ≡ d (mod n), then ac ≡ bd (mod n).

Homework Equations

Premise: a - b = n⋅x (1) and c - d = n⋅y (2), for some x, y ∈ Z.

Conclusion: ac - bd = n⋅z, where z ∈ Z and z = ___.

The Attempt at a Solution

Suppose a, b, c, d ∈ Z, a ≡ b (mod n) and c ≡ d (mod n), for which n ∈ N.
Hence n|(a-b) and n|(c-d). Further a - b = n⋅x and c - d = n⋅y, for some x, y ∈ Z.
...... I'm stuck.

I tried multiplying (1) and (2) but I got ac - bc - ab + bd = n^2 ⋅x ⋅y.
Addition won't work either. I think there must be a typographical error. Can you please tell if there is a solution to this problem?

 

[HallsofIvy]

Instead of "a- b=nx" and "c- d=ny" use the equivalent "a= b+ nx" and "c= d+ ny". Now, what is ac?

 

[RichardParker]

Thanks! So silly I haven't seen that