A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Empty

  • one year ago

Prove

  • This Question is Closed
  1. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    For a,b integers greater than 1, if \(\sqrt{2ab} \in \mathbb{Z}\) then \(a^2n^4+b^2\) is not prime for n>1.

  2. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I guess this is either going to be hard or easy I'm not sure which it is cause I know the answer lol

  3. amilapsn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    use \[a^2+b^2=(a+b)^2-2ab\]

  4. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yep that'll do it. :D

  5. amilapsn
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    :D

  6. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Clever!

  7. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I came up with this after reading this http://www.cut-the-knot.org/blue/McWarter.shtml to try to generalize this question at the end. :P

  8. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    spoiler : \(\color{white}{\sqrt{2ab} \in \mathbb{Z}\implies ab=2c^2 \\~\\a^2n^4+b^2 = (an^2)^2+b^2 = (an^2+b)^2-2an^2b = (an^2+b)^2-(2cn)^2}\)

  9. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeah that's exactly how I did, it, with c and everything haha.

  10. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    little challenge : prove that \(5\) is a divisor whenever \(5\nmid n\)

  11. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oooh interesting I'm thinking.

  12. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sry that is a blunder, doesn't work...

  13. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I am still interested in how you thought it would have worked though :D

  14. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    let \(a=2mx^2\) and \(b=my^2\) and let \(5\nmid n\), then : \[a^2n^4+b^2 = 4m^2x^4n^4+m^2y^4\equiv 4m^2+m^2\equiv 0\pmod{5}\]

  15. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    similarly a fake proof can be made for the other case : \(a=mx^2\) and \(b=2my^2\)

  16. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    btw \(m\) is the square free factor, that must match in both \(a\) and \(b\) for \(2ab\) to be a perfect square

  17. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Here is the revised challenge (not a challenge anymore) : prove that \(5\) is a divisor whenever \(5\nmid n,a,b\)

  18. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i lost interest immediately because of the constraint, \(5\nmid n,a,b\)

  19. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.