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

estudier Group TitleBest ResponseYou've already chosen the best response.1
Looks like a square in integers to me:)
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
OK, naturals if u insist
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
Number theory type question.....
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
3215 = 17 is not square
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
k=4,6 as of now
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
That's two....
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
next one seems a bit far
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
"a bit far" ?
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
N is a large place:)
 one year ago

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

estudier Group TitleBest ResponseYou've already chosen the best response.1
Hint Consider odd and even k separately
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
maybe 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..
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
the k which i used on RHS is different from 2^k
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
@shubhamsrg That looks nearly right for even k....
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.4
2^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
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.1
similar to what i was going to write..
 one year ago

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

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.4
for 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
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.4
again we get the same solutions for odd case k = {4, 6}
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.1
K, 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
 one year ago
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