Question

2^k -15 is square. Find all k (integers)

Answer 1

Does that mean \(2^k-15=a^2, a\in \mathbb N\)?

Answer 2

Looks like a square in integers to me:-)

Answer 3

OK, naturals if u insist

Answer 4

Number theory type question.....

Answer 5

k=5

Answer 6

32-15 = 17 is not square

Answer 7

k=4,6 as of now

Answer 8

That's two....

Answer 9

next one seems a bit far

Answer 10

"a bit far" ?

Answer 11

N is a large place:-)

Answer 12

If u are saying there are no more solutions, saying it is a bit far is insufficient evidence..:-)

Answer 13

Hint Consider odd and even k separately

Answer 14

maybe i got somewhere

2^k = m^2 or 2m^2

for 2^k = m^2

m^2 - 15 = k^2

(m+k)(m-k) = 1*3*5

on comparison,

m+k = 5 and m-k =3
=>m=4 -->1st solution

again on comparison,
m+k =15
m-k =1
m=8 -->2nd solution

just have to see through 2m^2 now..

Answer 15

the k which i used on RHS is different from 2^k

Answer 16

@shubhamsrg That looks nearly right for even k....

Answer 17

2^k-15 = p^2

2^k-p^2 = 15

lets say k is even, k = 2t
2^2t - p^2 = 15
(2^t + p)(2^t-p) = 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

Answer 18

similar to what i was going to write..

Answer 19

For the odd case, use the fact that any square divided by 3 leaves a remainder of 0 or 1

Answer 20

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

Answer 21

again we get the same solutions for odd case k = {4, 6}

Answer 22

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