Tuesday, 19 March 2019

proof explanation - If p1,...,pn are positive prime numbers then sqrtp1cdotspnnotinBbbQ




We want to proove that:




If p1,...,pn are positive, distinct, prime numbers then
p1pnQ.




Let's assume that p1pnQ. Then, (a,b)Z×Z:p1pn=ab with gcd(a,b)=1. So, a2=p1pnb2. But how do we continue? Is this technique right or should we follow something different?




PS: This is a part this proof, and I would like to discuss it.



Thank you.


Answer



The classic proof that 2Q readily extends to this case, to wit:



if



p1p2...pn=ab




with a,bN, gcd(a,b)=1, then



p1p2pnb2=a2,



whence p1a2; thus p1a and so a=p1c; thus



a2=p21c2=p1p2pnb2,



whence




p1c2=p2p3pnb2;



since the pi are distinct, we must have p1b2, whence p1b, contradicting our assumption that gcd(a,b)=1.



I think the ancient roots or this proof are worthy or respect, and I also note the method readily extends to show many similar propositions hold.



I presented this answer because it seems to me that the linked proof is pretty complex for this particular problem, although it is certainly engaging in and of itself, and leads in entaging directions.


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