Can anybody provide any information about this equation? $2^2 = \left(x+\frac{1}{x}\right)^2 - \left(x-\frac{1}{x}\right)^2 $

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Can anybody provide me with any information about this fascinating equation?

$$2^2 = \left(x+\frac{1}{x}\right)^2 - \left(x-\frac{1}{x}\right)^2 $$

I have been told many different things:

1) It is a mathematical Identity.

2) It is not a mathematical identity (because it does not work for zero).

3) It is a special case of the difference of squares equation.

Can anybody give me any concrete facts? The reason i ask is because it's so interesting. For example:

We can link this equation with Pythagoras’s theorem, so that for each number (x) we have a corresponding right angled

triangle and each triangle has height (A) = 2, base (B) = x-1/x and hypotenuse (C) = x+1/x.

triangle formation

As you can see, we can also link it with Trigonometry.

Triangle Formation Animation

And because tangents to an arc are always reciprocals, we can use it to get the following equations:

Tangent equations

SO, why does an equation this interesting not have a name (or does it?) and why is it not more well known?

I have been playing around with the geometry of this equation for years and it has allowed me to create many tools to help visualize the symmetry of a number or angle, for example:

Symmetry Animation

We can also link this equation with the metallic means.

Does this equation work for zero?

Assuming that (x) is always >= 1 and that (1/x) is always <= 1.

zero

As you can see, these variables always cancel each other out. Substituting 1/x = 0 (zero) and x = ∞ (infinity).

infinity zero

Tangents to an arc are always reciprocals.

Tangents to the arc

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The identity $$4xy=(x+y)^2-(x-y)^2$$ can be used to show that if two numbers have a fixed sum, their product is greatest when they are equal, as well as the AM/GM inequality for two numbers. It is well known and useful.

If we fix $xy=C$ then the identity $4C=(x+\frac Cx)^2-(x-\frac Cx)^2$ applies where all the terms are defined ie for $x\neq 0$ (note that $C$ need not be positive here).

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Using the formula $$a^2-b^2=(a-b)(a+b)$$ we get $$(x+\frac{1}{x}+x-\frac{1}{x})(x+\frac{1}{x}-x+\frac{1}{x})=2x\cdot \frac{2}{x}=4$$ if $$x\neq 0$$