Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

mukushla

for how many integers \(m\) the expression\[1+m+m^2+m^3+m^4\]becomes a perfect square?

  • 11 months ago
  • 11 months ago

  • This Question is Closed
  1. gorvs
    Best Response
    You've already chosen the best response.
    Medals 0

    1+m+m^2(1+m+m^2) \[1+m+m ^{2}(m+1)^{2}\]

    • 11 months ago
  2. gorvs
    Best Response
    You've already chosen the best response.
    Medals 0

    sry

    • 11 months ago
  3. gorvs
    Best Response
    You've already chosen the best response.
    Medals 0

    it will be perfect square for m=0 only

    • 11 months ago
  4. John_ES
    Best Response
    You've already chosen the best response.
    Medals 0

    Well, numerically I find others m not zero, m=-1,m=0 and m=3. A more rigorous proof is needed for all m in integers.

    • 11 months ago
  5. gorvs
    Best Response
    You've already chosen the best response.
    Medals 0

    1+m(m+1)+m^3(m+1)=1+(m+1)(m+m^3) =1+(m+1)(m^2(m+1)) =1+(m^2)*(m+1)^2 =1+(m(m+1))^2 except zero it will not be a perfect square bcoz after sqauring 1is added which makes it a perfect square

    • 11 months ago
  6. shubhamsrg
    Best Response
    You've already chosen the best response.
    Medals 1

    you have done a mistake in your 2nd step @gorvs

    • 11 months ago
  7. gorvs
    Best Response
    You've already chosen the best response.
    Medals 0

    sryyyyyyyyyyyy

    • 11 months ago
  8. Loser66
    Best Response
    You've already chosen the best response.
    Medals 0

    all you are off line, I can write whatever I like, I choose (m^2 +1) ^2 as my perfect square, so \[(m^2+1)^2=m^4+2m^2+1\] to find out the solution, I let \[m^4+2m^2+1=m^4+m^3+m^2+m+1\] so, when \[m^2=m+m^3\], I have perfect square \[m^2=m^2(\frac{1}{m}+m)\] ---> \[\frac{1}{m}+m=1\] \[m^2-m+1=0\]no real solution for this.

    • 11 months ago
  9. shubhamsrg
    Best Response
    You've already chosen the best response.
    Medals 1

    I got a method but not too sure if its ideal lets take 1+ m + m^2 + m^3 + m^4 = p^2 for some integer p then m(1+m) + m^3 (1+m) = (p+1) (p-1) m(1+m)(1+m^2) = (p+1)(p-1).1 now comparing will get our answers, but it'll be bit lengthy as too many permutations possible like this one yields m=3 and p=11 m(1+m) = 1+p (1+m^2) = p-1 like wise we will make all possible comparisons and try to find out integer solutions

    • 11 months ago
  10. Jonask
    Best Response
    You've already chosen the best response.
    Medals 0

    i asked you this before @mukushla

    • 11 months ago
  11. mukushla
    Best Response
    You've already chosen the best response.
    Medals 0

    really, i cant remember, nice one :)

    • 11 months ago
  12. Jonask
    Best Response
    You've already chosen the best response.
    Medals 0

    http://openstudy.com/users/jonask#/updates/50783e67e4b02f109be41dff

    • 11 months ago
  13. Jonask
    Best Response
    You've already chosen the best response.
    Medals 0

    @mukushla

    • 11 months ago
  14. mukushla
    Best Response
    You've already chosen the best response.
    Medals 0

    so do u know complete solution now?

    • 11 months ago
  15. Jonask
    Best Response
    You've already chosen the best response.
    Medals 0

    well somehow i gave up and jus considered the fact that it is between two perfect squares so m=3

    • 11 months ago
  16. mukushla
    Best Response
    You've already chosen the best response.
    Medals 0

    • 11 months ago
    1 Attachment
    • Attachments:

See more questions >>>

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.