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MC III Number Theory 1.14

 
 
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Re: MC III Number Theory 1.14
by John Lensmire - Friday, 15 March 2024, 9:41 AM
 

Let me give the idea and a quick hint here. (Note: The course recording has the answer to the question I'll ask in the hint below.)

This problem is related to Bezout's Theorem:

  • If gcd(a,b) = 1, then there exist m and n so that a*m + b*n = gcd(a,b).

In fact, it's exploring the converse that is relevant here:

  • If there exist m and n so that a*m + b*n = k, then k is a multiple of gcd(a, b).
  • Or really the special case of: if there exist m and n so that a*m + b*n = 1, then gcd(a,b) = 1.

Start first by making sure these make sense.

Now back to the problem: To show (a+b)/(c+d) is irreducible, we need to show gcd(a+b, c+d) = 1. This means we want to look for m and n so that (a+b)*m + (c+d)*n = 1. Hint: Nothing above has used the fact that ad-bc=1, so that is probably relevant here!

Hope this help!