A community for students.
Here's the question you clicked on:
 0 viewing
Empty
 one year ago
Prove
Empty
 one year ago
Prove

This Question is Closed

Empty
 one year ago
Best ResponseYou've already chosen the best response.1For a,b integers greater than 1, if \(\sqrt{2ab} \in \mathbb{Z}\) then \(a^2n^4+b^2\) is not prime for n>1.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I guess this is either going to be hard or easy I'm not sure which it is cause I know the answer lol

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.3use \[a^2+b^2=(a+b)^22ab\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I came up with this after reading this http://www.cuttheknot.org/blue/McWarter.shtml to try to generalize this question at the end. :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1spoiler : \(\color{white}{\sqrt{2ab} \in \mathbb{Z}\implies ab=2c^2 \\~\\a^2n^4+b^2 = (an^2)^2+b^2 = (an^2+b)^22an^2b = (an^2+b)^2(2cn)^2}\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah that's exactly how I did, it, with c and everything haha.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1little challenge : prove that \(5\) is a divisor whenever \(5\nmid n\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Oooh interesting I'm thinking.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1sry that is a blunder, doesn't work...

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I am still interested in how you thought it would have worked though :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1let \(a=2mx^2\) and \(b=my^2\) and let \(5\nmid n\), then : \[a^2n^4+b^2 = 4m^2x^4n^4+m^2y^4\equiv 4m^2+m^2\equiv 0\pmod{5}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1similarly a fake proof can be made for the other case : \(a=mx^2\) and \(b=2my^2\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1btw \(m\) is the square free factor, that must match in both \(a\) and \(b\) for \(2ab\) to be a perfect square

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Here is the revised challenge (not a challenge anymore) : prove that \(5\) is a divisor whenever \(5\nmid n,a,b\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i lost interest immediately because of the constraint, \(5\nmid n,a,b\)
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.