How can continued fractions be used to prove a deep result in mathematics?

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Homework Statement


Let x be any positive real number and suppose that ##x^2-ax-b=0## where ##a,b## are positive. I would like to use the equation that I provided in relevant equations which I proved to prove that
$$
\sqrt{\alpha^{2}+\beta}=\alpha+\cfrac{\beta}{2 \alpha+\cfrac{\beta}{2 \alpha+\cfrac{\beta}{2\alpha+\ddots}}}
$$
where ##\alpha,\beta>0##.

Homework Equations


I proved that
$$
x=a+\cfrac{b}{a+\cfrac{b}{a+\cfrac{b}{a+\ddots}}}.
$$

The Attempt at a Solution


I tried to do things like find values of ##a,b## so that when I transformed the equation ##x^{2}-ax-b=0## into a continued fraction that I would get the desired continued fraction with ##x=\sqrt{\alpha^{2}+\beta}## but that didn't work out.

I also tried changing ##\sqrt{\alpha^{2}+\beta}## directly into a continued fraction using the canonical continued fraction algorithm but I then had to consider different values of ##\beta## which would give me different continued fractions that I didn't really know how to combine to create the desired continued fraction.

I tried to plug ##x=\sqrt{\alpha^{2}+\beta}## into ##x^{2}-ax-b=0## and then solve for ##a,b## but that didn't get too far with two variables and one equation.
 
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If you set a = 2α and b = β, what are the roots of x2-ax-b?
 
haruspex said:
If you set a = 2α and b = β, what are the roots of x2-ax-b?

I don't know why but I remember trying that and it didn't work but now that I try it again after you mention it, it works. Thanks!
 
DeadOriginal said:
I don't know why but I remember trying that and it didn't work but now that I try it again after you mention it, it works. Thanks!
A deep result in maths is that arithmetic and algebraic results are NOT static over time, but depend on what help you get!
:smile:
 
arildno said:
A deep result in maths is that arithmetic and algebraic results are NOT static over time, but depend on what help you get!
:smile:

Amen! Haha.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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