## estudier 2^k -15 is square. Find all k (integers) one year ago one year ago

1. klimenkov

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

2. estudier

Looks like a square in integers to me:-)

3. estudier

OK, naturals if u insist

4. estudier

Number theory type question.....

5. sauravshakya

k=5

6. estudier

32-15 = 17 is not square

7. hartnn

k=4,6 as of now

8. estudier

That's two....

9. hartnn

next one seems a bit far

10. estudier

"a bit far" ?

11. estudier

N is a large place:-)

12. estudier

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

13. estudier

Hint Consider odd and even k separately

14. 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..

15. shubhamsrg

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

16. estudier

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

17. 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

18. shubhamsrg

similar to what i was going to write..

19. estudier

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

20. 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

21. ganeshie8

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

22. estudier

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

23. sauravshakya