Question #96099
Let a and b be integers. Prove that (a, b) = 1 if and only if (a + b, ab) = 1.
1
Expert's answer
2019-10-08T01:51:04-0400

1)Suppose that (a,b)=d>1(a,b)=d>1 . So aa and bb are divisible by dd. Then a+ba+b and abab are divisible by dd, that is (a+b,ab)(a+b,ab) is divisible by dd .

So if (a+b,ab)=1(a+b,ab)=1, then (a,b)=1(a,b)=1.

2)Suppose that (a+b,ab)(a+b,ab) is divisible by some prime pp. Then abab and a+ba+b are divisible by pp.

Since a+ba+b is divisible by pp, there is such kk that a+b=pka+b=pk.

Since abab is divisible by pp, we have that aa is divisible by pp or bb is divisible by pp.

If aa is divisible by pp, then a=pla=pl for some ll, then b=pka=p(kl)b=pk-a=p(k-l), that is bb is divisible by pp.

By symmetry we obtain that if bb is divisible by pp, then aa is divisible by pp.

In both cases aa and bb are divisible by pp, so (a,b)(a,b) is divisible by pp.

Therefore if (a,b)=1(a,b)=1, then (a+b,ab)=1(a+b,ab)=1.

From 1) and 2) we obtain that (a+b,ab)=1(a+b,ab)=1 if and only if (a,b)=1(a,b)=1.


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