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