anonymous
  • anonymous
2^k -15 is square. Find all k (integers)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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klimenkov
  • klimenkov
Does that mean \(2^k-15=a^2, a\in \mathbb N\)?
anonymous
  • anonymous
Looks like a square in integers to me:-)
anonymous
  • anonymous
OK, naturals if u insist

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anonymous
  • anonymous
Number theory type question.....
anonymous
  • anonymous
k=5
anonymous
  • anonymous
32-15 = 17 is not square
hartnn
  • hartnn
k=4,6 as of now
anonymous
  • anonymous
That's two....
hartnn
  • hartnn
next one seems a bit far
anonymous
  • anonymous
"a bit far" ?
anonymous
  • anonymous
N is a large place:-)
anonymous
  • anonymous
If u are saying there are no more solutions, saying it is a bit far is insufficient evidence..:-)
anonymous
  • anonymous
Hint Consider odd and even k separately
shubhamsrg
  • shubhamsrg
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..
shubhamsrg
  • shubhamsrg
the k which i used on RHS is different from 2^k
anonymous
  • anonymous
@shubhamsrg That looks nearly right for even k....
ganeshie8
  • ganeshie8
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
shubhamsrg
  • shubhamsrg
similar to what i was going to write..
anonymous
  • anonymous
For the odd case, use the fact that any square divided by 3 leaves a remainder of 0 or 1
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
again we get the same solutions for odd case k = {4, 6}
anonymous
  • anonymous
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
anonymous
  • anonymous
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