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M and n are not arbitrary real numbers, but arbitrary integers.

Otherwise, you can mention it.

counter-example

Let a=b=c= 1, m=0.5, n= 1.

If changed to:

Setting a

b

C is an arbitrary integer. When c | a and c | b, there is c|(ma+nb), where m and n are arbitrary integers.

This is an obvious conclusion.

But it is also possible to prove.

The following definitions are used:

If c|a, there must be an integer x that makes a=cx.

On the other hand, if there is an integer x that makes a=cx, then c|a

Then the proof is as follows:

C|a deduces that a=cx.

C|b deduces that b=cy.

Where x and y are integers.

Then ma+nb=mcx+ncy=c(mx+ny)

Where mx+ny is an integer.

By definition, c|(ma+nb)