This is another demonstration of a proof by the Polya method.
Proof.
Question 1. What things this statement is about?
Answer . This statement is about numbers m,a,b and two g.c.d.'s (a,b) and (ma,mb). Let us denote (a,b) by d and (ma,mb) by e. (You must always denote all things which the statement is dealing with.)
Question 2. Can we rewrite our statement using this new variables (d and e)?
Answer . Yes, we can. We have to prove that md=e.
Question 3. What is d?
Answer . d is the greatest common divisor of a and b.
Question 4. What does it mean?
Answer . This means two things:
a) d divides a and b.
b) d is the greatest number which divides both a and b.
Question 5. What is e?
Answer . e is the greatest common divisor of ma and mb.
Question 6. What does it mean?
Answer . This means two things:
a') e divides ma and mb.
b') e is the greatest number which divides both ma and mb.
Question 7. Can we rewrite the statement which we are proving using this new information?
Answer. Yes, we can. We have to prove that md=e which means that
(1) md divides ma and mb.
(2) md is the greatest number which divides both ma and mb.
Remark. An important thing happened: we divided our statement into two statements which seem to be easier to prove than the initial one.
Question 8. Can we prove one of these statements?
Answer . Yes, we can prove statement (1). Indeed, we know that d divides a. Therefore md divides ma. (In more details: d divides a means a=kd for some k. Then ma = mkd = k(md), whence md divides ma.) Similarly, d divides b whence md divides mb. Thus md divides both ma and mb, and statement (1) is proved.
Question 9. Can we prove statement (2)?
Answer . No, it does not seem easy to do.
Question 10. Can we rewrite our initial statement in another form using the information that we have got?
Answer . Yes, we have rewritten our statement using the information about e. Now we can use the information about d (we have not used the information that d is the g.c.d. of a and b). We have to prove that md=e which can be rewritten in the form e/m=d. This means that we have to prove the following three statements:
(3) e/m is an integer.
(4) e/m divides a and b.
(5) e/m is the greatest integer which divides both a and b.
Question 11. Can we prove (3)?
Answer . Yes, we can. Indeed, m divides ma and mb. Hence em by a property of greatest common divisors it divides the g.c.d. of ma and mb. Thus m divides e which means that e/m is an integer.
Question 12. Can we prove (4)?
Answer . Yes, we can. We know that e divides ma. This means that ma=ke for some k. Therefore a=ke/m = k(e/m), so e/m divides a.
Similarly, e/m divides b.
Question 13. Can we prove (5)?
Answer . No, it seems to be not so easy to do.
Question 14. We have proved a lot of things. Can we deduce our statement from the statements that we have proved?
Answer . Let us write down all the statement that we have proved.
a) d divides a and b.
b) d is the greatest number which divides both a and b.
a') e divides ma and mb.
b') e is the greatest number which divides both ma and mb.
(1) md divides ma and mb.
(3) e/m is an integer.
(4) e/m divides a and b.
Now let's see.
We have that d is the greatest number which divides both a and b. We have also that e/m divides both a and b. Hence e/m does not exceed d.
We have that e is the greatest number which divides both ma and mb, and we have that md divides both ma and mb. We can deduce that md does not exceed e.
The first of these inequalities may be rewritten in the form e does not exceed md. Therefore we have that md does not exceed e and e does not exceed md. This implies that md=e and we are done!!
The proof is complete.