Square root of a Mersenne Number is irrational

[1MileCrash]

Homework Statement

A user on math.se wanted to prove that any Mersenne number m = 2^n - 1 has an irrational square root for n > 1. So, it can be proved rather easily that any non-perfect square has an irrational root, and that all of the numbers to be considered are not perfect squares. However, the user requested a solution that did not use this idea, and that just focused on the Mersenne numbers themselves. I'm not sure why one would do this besides the challenge.

A few other posters proved this by using binary. I don't know anything about that. This discouraged me, as I submitted a simple proof by contradiction using just basic algebra. Can anyone tell me if my proof actually works?

Homework Equations

The Attempt at a Solution

Assume by way of contradiction that there exist integers a, b, such that gcd(a,b) = 1 and (a/b) = sqrt(2^n - 1).

Then
\frac{a^{2}}{b^{2}} = 2^{n} - 1
\frac{a^{2}}{b^{2}} + 1 = 2^{n}

Since 2^n is even, a^2/b^2 is odd.
Since the quotient a^2/b^2 is odd, either both a^2 and b^2 are even or both a^2 and b^2 are odd.

Case #1; a^2 and b^2 are both even.

Then, a is even and b is even.
Contradiction, gcd(a,b) = 1.

Case #2; a^2 and b^2 are both odd.

Then, a is odd and b is odd.
Let a = 2x + 1 and b = 2y + 1

\frac{a^{2}}{b^{2}} + 1 = 2^{n}

a^{2} + b^{2} = 2^{n}b^{2}

(2x + 1)^{2} + (2y + 1)^{2} = 2^{n}b^{2}

4x^{2} + 4x + 4y^{2} + 4y + 2 = 2^{n}b^{2}

2(2x^{2} + 2x + 2y^{2} + 2y + 1) = 2^{n}b^{2}

2x^{2} + 2x + 2y^{2} + 2y + 1 = 2^{n-1}b^{2}

2(x^{2} + x + y^{2} + y) + 1 = 2^{n-1}b^{2}

Contradiction; 2^(n-1) is even since n > 1, and b^2 is odd. Therefore since an odd times an even is always even, the RHS is even. The LHS is odd by definition.

The shows that a contradiction occurs in all cases, so there is no such a, b.

 

[jbunniii]

Your proof looks valid to me.

 

[1MileCrash]

Thanks for the response. It was pretty fun to write, I'm glad it's ok.