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Question

this is a group related question.

we let x,y,z belong to N, if (x, y) = 1 and x | (yz), how to prove x|z?

note "|" sign is not division sign!

Answer 1

If x and y are relatively prime, there exist integers s and t such that,
s*x + t*y = 1,

multiply both sides by z, giving
z*s*x + z*t*y = z

But x obviously divides z*s*x, and we assumed x divides y*z so it divides z*t*y, so x divides the sum z*s*x + z*t*y = z

Answer 2

Since x is a factor of yz, then there exists an integer, r, such that rx = yz. Also since x and y are relatively prime, there exist integers, m and n, such that mx + ny = 1.

So mx + ny =1, then mxz + nyz = z (just muiltiplied everything by z).
But remember, yz = rx, so substitute this in:
Now mxz +nyz = z becomes mxz + nrx = z. Now factor, and you get z = x(mz + nr).
Therefore, x divides z.