A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
estudier
 2 years ago
2^k 15 is square. Find all k (integers)
estudier
 2 years ago
2^k 15 is square. Find all k (integers)

This Question is Closed

klimenkov
 2 years ago
Best ResponseYou've already chosen the best response.0Does that mean \(2^k15=a^2, a\in \mathbb N\)?

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Looks like a square in integers to me:)

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1OK, naturals if u insist

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Number theory type question.....

estudier
 2 years ago
Best ResponseYou've already chosen the best response.13215 = 17 is not square

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.0next one seems a bit far

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1If u are saying there are no more solutions, saying it is a bit far is insufficient evidence..:)

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Hint Consider odd and even k separately

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.1maybe i got somewhere 2^k = m^2 or 2m^2 for 2^k = m^2 m^2  15 = k^2 (m+k)(mk) = 1*3*5 on comparison, m+k = 5 and mk =3 =>m=4 >1st solution again on comparison, m+k =15 mk =1 m=8 >2nd solution just have to see through 2m^2 now..

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.1the k which i used on RHS is different from 2^k

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1@shubhamsrg That looks nearly right for even k....

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.42^k15 = p^2 2^kp^2 = 15 lets say k is even, k = 2t 2^2t  p^2 = 15 (2^t + p)(2^tp) = 15 solving, we get few solutions lets say k is off, k = 2t+1 2^2t+1  p^2 = 15 (2^2t+1/2 + p)(2^2t+1/2  p) = 15 solving we may not get any solutions on cursory check.. only k = 4,6 are the solutions

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.1similar to what i was going to write..

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1For the odd case, use the fact that any square divided by 3 leaves a remainder of 0 or 1

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.4for the even case, k = {4,6} for the odd case : 2.2^2t+1/2 = 16 2.2^2t+1/2 = 8 we need to solve above both

ganeshie8
 2 years ago
Best ResponseYou've already chosen the best response.4again we get the same solutions for odd case k = {4, 6}

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1K, for the odd you can also say that 2^k has remainder 2 on division by 3 and then so does 2^k 15 which is not 0 or 1 so there are no solutions when k is odd
Ask your own question
Ask a QuestionFind 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.