SOLUTION: Find all the positive integers k for which 7×(2^k)+ 1 is a perfect square. I started by setting it to 7×(2^k)+1=x^2 where x^2 represents perfect squares. Then i rearranged to 7

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Find all the positive integers k for which 7×(2^k)+ 1 is a perfect square. I started by setting it to 7×(2^k)+1=x^2 where x^2 represents perfect squares. Then i rearranged to 7      Log On


   

Question 1122720: Find all the positive integers k for which 7×(2^k)+ 1 is a perfect square.
I started by setting it to 7×(2^k)+1=x^2 where x^2 represents perfect squares. Then i rearranged to 7×(2^k)=x^2-1. Then I used difference of two squares to get 7×(2^k)=(x+1)(x-1). I then used logs to rearrange it for k, but I am not sure if that is right or if I am even on the right path.
Thank you for taking the time to attempt this in advance!!
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Find all the positive integers k for which 7×(2^k)+ 1 is a perfect square.
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The only integer value for k which has a perfect square is 5
I put this equation into my Ti 83, and checked the table, (x=k)
y = sqrt%287%282%5Ex%29%2B1%29
x=5 is only value that produces an integer for y=15, (225 is a perfect square)
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However, pursuing your method
rearrange to:
%28x%5E2-1%29%2F7 = 2^k
then try various values for K, say k=2
= %28x%5E2-1%29%2F7 = 4
x^2 - 1 = 28
x^2 = 29 obviously no perfect square there:
:
try k=3,4, and 5
k=5
%28x%5E2-1%29%2F7 = 2^5
x^2 - 1 = 32*7
x^2 = 224 + 1
x^2 = 225 a perfect square
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I think that is the only one, but you can check with other values for k yourself
:
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