Question 1122720
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(7(2^x)+1)}}}
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:
 {{{(x^2-1)/7}}} = 2^k
then try various values for K, say k=2
 = {{{(x^2-1)/7}}} = 4
x^2 - 1 = 28
x^2 = 29 obviously no perfect square there:
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try k=3,4, and 5
k=5
{{{(x^2-1)/7}}} = 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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