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

Mathematics
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Does that mean \(2^k-15=a^2, a\in \mathbb N\)?
Looks like a square in integers to me:-)
OK, naturals if u insist

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Other answers:

Number theory type question.....
k=5
32-15 = 17 is not square
k=4,6 as of now
That's two....
next one seems a bit far
"a bit far" ?
N is a large place:-)
If u are saying there are no more solutions, saying it is a bit far is insufficient evidence..:-)
Hint Consider odd and even k separately
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..
the k which i used on RHS is different from 2^k
@shubhamsrg That looks nearly right for even k....
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
similar to what i was going to write..
For the odd case, use the fact that any square divided by 3 leaves a remainder of 0 or 1
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
again we get the same solutions for odd case k = {4, 6}
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
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