# Prerequisites

To understand this post, you’ll need to know and A good grasp on Algebra is also nice. This might be helpful, but knowing these are not necessary.

# Preamble

One morning I was thinking of the following configuration:

Consider with and Then for all construct such that

It doesn’t take long to figure out that this is very easy to express with polar coordinates (or complex, whatever flavor you desire). If and then in polar coordinates, where

Then the realization strikes; **none of the triangles “overlap” each other!** In other words, there is no pair of integers such that are collinear. What does this mean? Well, there are no integers such that for some integer This basically means that there is no rational number such that

# The Meat

We claim that is irrational. **This is the same as claiming ** **is irrational.**

Suppose otherwise. Then there must be some positive integer such that But notice that is a polynomial about Similarly, is a polynomial about

We claim that for all Adding and expanding out using the sum-difference identities yields which is true.

Now we have a recurrence for This general recurrence is called the **Chebyshev Polynomial of the First Kind**. In fact, generally, the Chebyshev Polynomial of the First Kind states the following:

Now we have to prove that is not a root of **any** of these polynomials. What follows is probably the most ingenious yet simplest step of this problem (and the extension, Niven’s Theorem) – an application of the Rational Root Theorem. We claim that the coefficient of the leading term of is The proof is fortunately very obvious; we can discard the and notice that the leading coefficient of is multiplied by By the Rational Root Theorem, **CANNOT** be a root of for **ANY** because **NEVER** divides Isn’t it beautiful?

# Generalization

Now let’s talk about Niven’s Theorem. It states that the only non-negative values of such that is rational are To do this, we need to expand the Chebyshev Polynomial of the First Kind a little bit more. The Rational Root Theorem will be more powerful if we have the leading coefficient and the constant value; let’s try to look for a pattern of the constant value.

Actually, let’s make it a bit more obvious.

Again, this is very easy to prove using induction. So what does this mean? By the Rational Root Theorem, the only possible values of are either or of the form for even of and for odd we can also have things of the form (Remember is non-negative; negative values are symmetric anyways.) So now all that remains is to prove that these are not roots.

To utilize modular arithmetic, we’ll divide by for $P_n(x).$ If we let we are basically “reversing” the polynomial. Let’s rewrite the polynomials in hope we see a pattern:

Let Notice the recursion is We’ll write the first few

The only possible roots are

Notice that for odd the leading coefficient is This then causes an obvious problem $\pmod n,$ so it’s proven for odd (It is obvious that this is impossible if or because we need everything to be an integer if we are to get rid of the constant.)

For even notice that the constant term is of the form while the terms with a coefficient of are of the form $(2^{k})^{2j}),$ so it won’t work out either. Thus we have proven this is impossible for all rational numbers except for our three exceptions. (For is an exception.)

And that’s Niven’s Theorem!

# References

Stack Exchange: Is arccos 3/5 rational?

Stack Exchange: 1/pi arccos (1/3) is irrational.

Wikipedia: Chebyshev Polynomials

Wolfram Mathworld: Chebyshev Polynomials of the First Kind

Brilliant: Chebyshev Polynomials