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Another 2007 Problem (Posted on 2007-02-19) Difficulty: 3 of 5
Analytically determine all possible integer solutions (x,y) of each of the following equations:

(A) √(x+ √(x+ √( x +......+ √(x + √x))))) = y

(B) √(x+ √(x+ √(x + √(x ......+ √(x + √(2x))))) = y, with x < 180,000

The square root symbol in the each of the above relationships is repeated 2007 times.

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Solution to A | Comment 2 of 7 |

For equation A, there is only one solution as follows,

The only hope for having y to be an integer is that after each square root is taken, the result is an real integer, since adding an irrational number to a rational number must result in an irrational number, and thus each additional x and square root will keep it irrational, which will result in an irrational y and not an integer. Similarly, the square root of a complex number must be complex, so y will be complex and not real. So we can conclude x and y must be non-negative

However, sqrt(x+sqrt(x))=z has only one integer solution for x,z, x=z=0. To see this, square both sides, to give x+sqrt(x)=z and sqrt(x)(sqrt(x)+1)=z. It is clear here that x must be a perfect square, or z would be irrational. So sqrt(x) and sqrt(x)+1 are integers, but if sqrt(x)>1, then they are relatively prime. So they must both be perfect squares, but it is easily seen that there are no consecutive perfect squares above 1. So x must be 1 or 0. If x=1, z=sqrt(2), which is irrational, but if x=0, z=0, which works.

Since sqrt(x+sqrt(x)) is irrational for nonzero integer x, then in equation A, y will be irrational for nonzero integer x. So for the first equation, x=0, y=0 is the only solution.


  Posted by Gamer on 2007-02-19 19:41:33
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