# Proof by contradiction - polynomials and infinite primes

by bloynoys
Tags: contradiction, infinite, polynomials, primes, proof
 P: 24 1. The problem statement, all variables and given/known data Two Questions: 1. Prove, by contradiction, that if a and b are integers and b is odd,, then -1 is not a root of f(x)= ax^2+bx+a. 2. Prove, by contradiction, that there are infinitely many primes as follows. Assume that there only finite primes. Let P be the largest prime. Explain why there is a prime dividing P!+1 and find the the contradiction. 2. Relevant equations For both, assume the contradiction work towards finding it is impossible. 3. The attempt at a solution 1. (x-1)(x-a)=ax^2+bx+a Not sure where to go from there 2. This is not a normal infinite prime solutions as we have gone over a few of the solutions in class. I am not sure what she means by "there is a prime dividing P!+1" as that isn't really a clear sentence. Any Ideas?
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 Quote by bloynoys 1. The problem statement, all variables and given/known data Two Questions: 1. Prove, by contradiction, that if a and b are integers and b is odd,, then -1 is not a root of f(x)= ax^2+bx+a. 2. Prove, by contradiction, that there are infinitely many primes as follows. Assume that there only finite primes. Let P be the largest prime. Explain why there is a prime dividing P!+1 and find the the contradiction. 2. Relevant equations For both, assume the contradiction work towards finding it is impossible. 3. The attempt at a solution 1. (x-1)(x-a)=ax^2+bx+a Not sure where to go from there
What does it mean for -1 to be a root of f?? What is the definition??

 2. This is not a normal infinite prime solutions as we have gone over a few of the solutions in class. I am not sure what she means by "there is a prime dividing P!+1" as that isn't really a clear sentence. Any Ideas?
What's wrong with that sentence?? "There is a prime number that divides P!+1" is a perfectly good sentence...
 P: 24 Maybe I am just confused to how to relate the P!+1 to P. Is it P/(P!+1)? Or (P!+1)/P and that just eliminates the "largest prime?" So it is the second largest prime plus one? And for the first one, a root means that it is one of the solutions to an equation. I realize this looks dumb as this is an easyish problem, but we are working on this in a high level undergrad math course. I am just drawing a blank on the intermediate steps to set these problems up. Oh and I believe I made a mistake. It should read: (x+1)(x+a)=ax^2+bx+a correct?
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Emeritus
P: 15,671

## Proof by contradiction - polynomials and infinite primes

 Quote by bloynoys Maybe I am just confused to how to relate the P!+1 to P. Is it P/(P!+1)? Or (P!+1)/P and that just eliminates the "largest prime?" So it is the second largest prime plus one?
You're making this far too difficult. For example, let P=3, then P!+1=7. So a prime that divides P!+1 is 7.

In general, there is always a prime that divides P!+1 for any P.

 And for the first one, a root means that it is one of the solutions to an equation. I realize this looks dumb as this is an easyish problem, but we are working on this in a high level undergrad math course. I am just drawing a blank on the intermediate steps to set these problems up. Oh and I believe I made a mistake. It should read: (x+1)(x+a)=ax^2+bx+a correct?
-1 being a root means that f(-1)=0 right? Now write out the definition of f and see what it gives you.
 P: 24 Wow, holy cow, that was a whole lot easier than I thought for the first one. I just had my head focused in on a multiple root function and didn't think about the f(x)=0. Thank you for that! And wow on the second one as well. She basically spelled out the definition of prime numbers, and then added one. It was the word dividing that had me confused. I knew they weren't as hard as I was making them but wow, I really shouldn't overlook stuff like that. Thank you for all your help!

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