Question #96782
Let 'a' and 'b' be relatively prime integers. Find all values of (a + 2b, a − 2b).
1
Expert's answer
2019-10-18T09:24:27-0400

Given that aa and bb are relatively prime integers. Then gcd(a, b)=1gcd(a,\ b)=1 or in equivalent notation (a, b)=1.(a,\ b)=1.

Let d=(a+2b, a2b).d=(a+2b, \ a-2b). Then it follows that d(a+2b)d|(a+2b) and d(a2b).d|(a-2b).

That is d(m(a+2b)+n(a2b)),d|(m(a+2b)+n(a-2b)),

2(a+2b)+2(a2b)=4a2(a+2b)+2(a-2b)=4a and thus d4a,d|4a,

2(a+2b)2(a2b)=4b2(a+2b)-2(a-2b)=4b and thus d4b.d|4b.

Hence d(4a, 4b)d|(4a,\ 4b). We have that d4.d|4.

We can rule out every possibility except d=1,2d=1,2 or d=4.d=4. However, we do not yet know which of those three values are actually possible.

Let's show that all possible values of d{1,2,4}d\in \lbrace 1,2,4\rbrace actually occur:

a=1,b=0a=1,b=0 gives d=1,d=1,

a=0,b=1a=0, b=1 gives d=2,d=2,

a=2,b=1a=2, b=1 gives d=4.d=4.

Therefore, the answer is d{1,2,4}.d\in \lbrace 1,2,4\rbrace.


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