What is the value of $$\frac{294395}{294393}-\left(\frac{588789}{588787}\right)^2?$$
Motivation
Have you ever tried pressing the square root button repeatedly on your scientific calculator to see how quickly (or slowly) a number reached $1$? If you were lucky, you may have seen that during this process, fractions come up on the screen of your calculator. It is famously known that the square root of any non-square positive integer is irrational, so it must be strange to see that rational values appear. The reason: most scientific calculators only evaluate arithmetically up to $12$ decimal places.
I came across the integer $1539929$ when trying out the RanInt#( function on my CASIO, with limits at $1$ and $10$ million. I then decided to perform the process described above. The first few iterations are as follows: $$1240.93..., 35.22, ..., 5.93..., ..., 1.003484..., ..., 1.000217..., 1.000108..., \frac{147201}{147193}, \frac{147199}{147195}, \frac{73599}{73598}, \frac{294395}{294393}, \frac{588789}{588787},...$$
Using the last two fractions as an example, I tried to find out by how much the calculator was out. That is, does $$\frac{294395}{294393}-\left(\frac{588789}{588787}\right)^2=0?\tag{1}$$
Proof that $(1)$ is false
If you insert the LHS of $(1)$ into your calculator, it would return the value $0$. This is not actually the case.
Suppose that $(1)$ is true. Then $294395(588787^2)=294393(588789^2)$.
Now $588787^2$ ends in $9$, so LHS ends in $5$, and $588789^2$ ends in $1$, so RHS ends in $3$.
We have reached a contradiction, since no integer ending in $5$ can end in $3$. This disproves $(1)$.
What is the value of the LHS of $(1)$?
Let $x=588789$ and $y=294395$. Then we can write the LHS as $$\frac{y}{y-2}-\left(\frac{x}{x-2}\right)^2$$ Call this $z$. Furthermore, $x=2y-1$, so $$z=\frac{y}{y-2}-\left(\frac{2y-1}{2y-3}\right)^2=\frac{y(2y-3)^2-(y-2)(2y-1)^2}{(y-2)(2y-3)^2}=\frac2{(y-2)(2y-3)^2}$$ Now use your calculator to plug in the value of $y$. You get $$\frac{294395}{294393}-\left(\frac{588789}{588787}\right)^2=1.959...\times 10^{-17}=1.96\times 10^{-17}$$ to $3$.s.f.
Generalisation
Consider the expression $$z=\frac{y}{y-a}-\left(\frac{cy-d}{cy-d-b}\right)^2$$ where $a,b,c,d$ are positive integers. The calculations above are very useful when $ac=2b$ and $b=2d$ (check!) What is $z$ in this case? Surprisingly, it turns out that it is independent of $b,c$ and $d$: $$z=\frac{a^3}{(y-a)(4y-3a)^2}$$
Example of generalisation
When $x=1120416$ and $y=373473$. This can 't be done using a calculator. But using the above method, we find that $$z=7.69\times 10^{-17}$$ to $3$.s.f.
There are of course many more generalisations that could be made (for instance, cube roots instead fo square roots). But I'll leave these for you to think about. :)
You've got a pair of numbers around $3\times10^5$ and a pair of numbers around $6\times10^5$. At a glance, you can see that $(3\times10^5)\times2=(6\times10^5)$ so we can simplify the problem somewhat.
Let $x=294395$. Then $2x=588790$ and the problem becomes:
$$\frac{294395}{294393}-\left(\frac{588789}{588787}\right)^2 \\ =\frac{x}{x-2}-\left(\frac{2x-1}{2x-3}\right)^2 \\ =\frac{x(2x-3)^2-(x-2)(2x-1)^2}{(x-2)(2x-3)^2} \\ =\frac{2}{(x-2)(2x-3)^2} $$
Can you take it from here?