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irreducible? (Posted on 2013-03-29) Difficulty: 2 of 5
Prove that the fraction (21n+4)/(14n+3) is irreducible for every natural number n.

Source: Russian IMO

No Solution Yet Submitted by Ady TZIDON    
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Solution Relatively easy (spoiler) | Comment 1 of 6
Useful lemma:  a and b are relatively prime if and only if (a-b) and b are relatively prime.

Well, (21n+4)/(14n+3) is irreducible if (21 n+4) and (14n+3) are relatively prime.

And this is the case if (7n +1) and (14n+3) are relatively prime.

And this is the case if (7n +1) and (7n+2) are relatively prime.

And this is the case if (7n +1) and 1 are relatively prime, which they obviously are for all natural n.

Therefore,  (21n+4)/(14n+3) is irreducible for all natural n

 



  Posted by Steve Herman on 2013-03-29 10:04:01
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