general relativity - why do x Schwarzschild radii equal time dilation effects of speed of light going y times faster than an object^2?

let me walk you through the math.

First you start with the gravitational time dilation formula where:

$$T_1=T\sqrt{1-\frac{2GM}{rc^2}}$$

and rather than entering $r$ for the radius we replace $r$ with the Schwarzschild radius formula $(2GM/c^2)x$ with an $x$ at the end representing how many Schwarzschild radii you are away from the center. This brings the formula to look like:

$$T_1=T\sqrt{1-\frac{2GM}{\frac{2GM}{c^2}xc^2)}}$$

Which when simplified breaks down to:

$$T_1=T\sqrt{1-\frac{1}{x}}$$

and if you make $T=1$ then you just get

$$T_1=\sqrt{1-\frac{1}{x}}$$

This is very similar to the one in many physics books $=\sqrt{1-r_0/r}$, where $r_0$ is equal to the Schwarzschild radius and then $r$ equals the radius from the center. The formula above it just makes it slightly simpler due to making $r_0$ equal to 1 and $x$ equal to how many radii a point you are observing is from the center of the mass.

That is the gravitational time dilation side portion of this relationship. Now for the velocity time dilation side we use a similar methodology and start with:

$$T_0=T\sqrt{1-\frac{v^2}{c^2}}$$

Now we make $T$ equal to 1, $v$ equal to one, and $c$ to $y$ because now we are going to make $c$ a variable.

$$T_0=\sqrt{1-\frac{1}{y^2}}$$

What you see now "$1/y^2$" is showing the velocity as a constant 1 and $y$ represents how much faster light is going than the velocity constant of 1. If the above were to show the fraction as $1/5^2$ then this would be the same as saying an object is going at a velocity 1/5th the velocity of light. So now if we solve the velocity and gravitational time dilation formulas so that we can see how they dilate time to come up with the same result:

$$\sqrt{1-\frac{1}{x}}=\sqrt{1-\frac{1}{y^2}}$$

We can simplify this to

$$x=y^2$$

What does this mean?