Relatively Prime Numbers proof

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To prove that a and b are relatively prime, it is shown that if (a, b) = 1, then a+b and b are also relatively prime, as well as a and a+b. The discussion highlights the need to demonstrate that if n divides both a+b and b, then n must equal ±1, confirming their relative primality. A proof by cases is suggested, starting with the assumption that n is a common factor of both a+b and b. The conclusion drawn is that since a and b share no common factors other than 1, it follows that a+b must also be relatively prime to both a and b. This establishes the equivalence of the three conditions regarding relative primality.
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1. Suppose that a and b are positive integers. Show that the following are equivalent: 1) a and b are relatively prime 2) a+b and b are relatively prime 3) a and a+b are relatively prime.



2. I know that for a and b to be relatively prime, (a,b) = 1 (that is, their greatest common divisor is 1). Or, there exists an integer n such that if n divides a and n divides b, then n = +/-1.



3. I'm starting by assuming that a and b are relatively prime. My problem is I don't know how to go about showing a+b and b are relatively prime (if I know how to show that, then "a and a+b are relatively prime" follows). This must be proof by cases, but where might I begin?
 
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Hint: supppose n is a factor of (a+b) and b.

Then b = nx and a+b = ny for some integers x and y

What does that tell you about the factors of a?
 
The factors of a are n and y-x. This means that a, b, and a+b all have a common factor. Since a and b are relatively prime, the common factor must be 1.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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