# What is $t_k$ in this context?

I am reading this paper which provides an argument for factoring $\sinh$ and $\sin$ using infinitesimals as Euler did but in a more rigorous way.

On the bottom of the page labeled 66 it states:

1. Euler uses the possibility (established in 151 of [1]) of factorizing $a^{\omega} - b^{\omega}$ [$\omega$ represents an infinitely large value] into factors. $$t_k = a^2 + b^2 - 2ab\cos\left(\frac{2\pi k}{\omega}\right)$$ where $0 < 2k < \omega$, to which we add a factor of $a-b$, and if $\omega$ is an even number, a factor of $a + b$ also.

I am not exactly sure what $t_k$ means in this situation. I analyzed a few values of $a^n - b^n$ to see if I could gather any idea of what was going on:

$$a^2 - b^2 = (a - b)(a + b)$$ $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ $$a^4 - b^4 = (a - b)(a + b)(a^2 + b^2)$$ $$a^5 - b^5 = (a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4)$$

So I can see that there is this idea that for even powers there is an $(a + b)$ term, but I don't see the relation to the rest of the stuff. I also tried finding an English translation of Euler's work, but could only find French and German translations. Can anyone provide any further elaboration on what $t_k$ is and how $a^n - b^n$ relates to it?

Well, two examples of $t_k$ are present in the factorizations you have shown. First, for $\omega=3$, $t_1$ is just your second factor $a^2+b^2+ab$ in the factorization of $a^3-b^3$, since $\cos(2\pi/3)=-1/2$. Similarly, for $\omega=4$, $t_1$ is $a^2+b^2$ since $\cos(2\pi/4)=0$. Your factorization of $a^5-b^5$ doesn't have any $t_k$ factors, but that's because you haven't factored it completely. If you factored that quartic $a^4-a^3b+a^2b^2+ab^3+b^4$, you could find that it factors as a product of two quadratics, which are just $t_1$ and $t_2$ (though admittedly it is not obvious how to come up with this factorization).
To understand what's going on in general, it is very helpful to first factor things over the complex numbers. Over the complex numbers, we can factor $a^n-b^n$ all the way down to linear factors: $$a^n-b^n=\prod_{m=0}^{n-1}(a-e^{2\pi i m/n}b).$$ Note that this just comes from the factorization $$x^n-1=\prod_{m=0}^{n-1}(x-e^{2\pi i m/n})$$ since the numbers $e^{2\pi i m/n}$ are the roots of the polynomial. Letting $x=a/b$ and multiplying by $b^n$, you obtain the factorization of $a^n-b^n$ above.
Now the idea is to pair up the complex conjugate factors in this product. For each $k$, $e^{2\pi i k/n}$ and $e^{2\pi i (n-k)/n}=e^{2\pi i n/n}\cdot e^{-2\pi i k/n}=e^{-2\pi i k/n}$ are complex conjugates. When we combine our factors for $m=k$ and $m=n-k$, we get $$(a-e^{2\pi i k/n}b)(a-e^{-2\pi i k/n}b)=a^2+b^2-(e^{2\pi i k/n}+e^{-2\pi i k/n})ab.$$ Since $e^{2\pi i k/n}+e^{-2\pi i k/n}=2\cos(2\pi k/n)$, this is exactly $t_k$.
So for each $k$ such that $0<2k<n$, we get a factor of $t_k$ in $a^n-b^n$ by pairing together the factor with $m=k$ and the factor with $m=n-k$. That leaves the factor with $m=0$ and (if $n$ is even) the factor with $m=n/2$ (which we can't pair up since it would pair up with itself). These are exactly the remaining factors $a-b$ and $a+b$.
Euler's factorization is then just the same thing, but with $n$ replaced by an "infinitely large integer" $\omega$.