# Methods for proving a function outputs an infinite number of integers

I have a function involving polynomials and the centre of the Binomial Triangle and I'd like to prove that the function produces a positive integer infinitely many times. I don't have any interest in what values the integers take so much, merely that they exist.

The function I have is:

$$f(n) = \dfrac{15n^5+23n^4-20n^3-56n^2-48n-16}{n^3(n+1)(n+2)^4}\begin{pmatrix}2n \\ n \end{pmatrix} + 4\dfrac{3n^2+6n+4}{n^3 (n+2)^3}$$

The only method I have thought of is that I could try to prove that $$\sin\left(\pi f(n)\right)$$ has an infinte number of roots. But that feels like kicking the can down the road as I don't know how to do that either!

Thank you for any and all help. Ben

Using partial fractions I get:

$$f(n) = \frac{\left( 4n^3+28n^2+84n+76 \right) \begin{pmatrix} 2n \\ n \end{pmatrix} - 2n-4}{(n+2)^4} - \frac{\begin{pmatrix} 2n \\ n \end{pmatrix} - 2}{n^3} - \frac{4\begin{pmatrix} 2n \\ n \end{pmatrix}}{n+1}$$

From here I can find conditions for each fraction separately.

$$\frac{\begin{pmatrix} 2n \\ n \end{pmatrix} - 2}{n^3} \in \mathbb{N} \implies n \text{ is prime > 3 (I found this on A000984)}$$

$$\frac{4\begin{pmatrix} 2n \\ n \end{pmatrix}}{n+1} \in \mathbb{N} \implies n \in \mathbb{N} \text{ (Always true)}$$

$$f(n) = \frac{\left( 4n^3+28n^2+84n+76 \right) \begin{pmatrix} 2n \\ n \end{pmatrix} - 2n-4}{(n+2)^4} - \color{Blue}{\frac{\begin{pmatrix} 2n \\ n \end{pmatrix} - 2}{n^3} - \frac{4\begin{pmatrix} 2n \\ n \end{pmatrix}}{n+1}}$$

This leaves the final condition:

$$\frac{\left( 4n^3+28n^2+84n+76 \right) \begin{pmatrix} 2n \\ n \end{pmatrix} - 2n-4}{(n+2)^4} \in \mathbb{N} \\ \implies \left( 4n^3+28n^2+84n+76 \right) \begin{pmatrix} 2n \\ n \end{pmatrix} \equiv 2n+4\pmod{n^4+8n^3+24n^2+32n+16}$$

This seems to hold true when $$n+2$$ is a prime.

So if this function outputs an infinite number of integers, the twin-prime conjecture should be true. If $$n$$ is the lower of a pair of twin primes, then $$f(n)$$ is an integer.

• I would start by rewriting those rational functions using partial fractions, to isolate the divisibility condition you need for each separate factor in the denominator. – Eric Wofsey Jan 1 at 22:38
• I would expect the product of the fraction and the binomial coefficient to be an integer almost always, i.e. for all but finitely many values of $n$, whereas the other fraction can clearly only be an integer for finitely many $n$. – Servaes Jan 1 at 22:38
• @Servaes: The first term is never an integer if $n$ is an odd prime (there is nothing to cancel the $n^3$ in the denominator). – Eric Wofsey Jan 1 at 22:40
• @EricWofsey Thanks Eric, I'll give that a go. – Ben Crossley Jan 1 at 22:51
• What makes you think that what you are trying to prove is true? Do you have any numerical evidence? – Somos Jan 2 at 0:30

## 1 Answer

Here is a proof that if both $$n$$ and $$n+2$$ are prime ($$n>3$$), then the output is integer: clearly it suffices to show that if $$n+2$$ is prime then $$\left(4n^3+28n^2+84n+76\right)\binom{2n}{n} \equiv 2(n+2) \bmod{(n+2)^4}.$$ Let $$p=n+2$$ be prime, then this is equivalent to showing that: $$\tag1 2\left(p^3+p^2+5p-3\right)\binom{2p-4}{p-2} \equiv p \bmod{p^4}.$$

Proof

Since $$n>3$$, $$p>3$$ and by Wolstenhome's theorem we have: \begin{align} \binom{2p-1}{p-1}&\equiv 1 \bmod p^3\\ p\binom{2p-1}{p-1}&\equiv p \bmod p^4\\ p\frac{2p-1}{p-1}\frac{2p-2}{p-2}\frac{2p-3}{p-3}\binom{2p-4}{p-4}&\equiv p \bmod p^4. \end{align}

But $$\binom{2p-4}{p-2}=\binom{2p-4}{p-4}\frac{p-1}{p-3}\frac{p}{p-2}.$$ Then \begin{align} p\frac{2p-1}{p-1}\frac{2p-2}{p-2}\frac{2p-3}{p-3}\frac{p-3}{p-1}\frac{p-2}{p}\binom{2p-4}{p-2}&\equiv p \bmod p^4 \end{align} That is \begin{align} \frac{(2p-1)(2p-2)(2p-3)}{(p-1)^2}\binom{2p-4}{p-2}&\equiv p \bmod p^4 \end{align}

But by doing the power expansion of $$\frac{1}{(p-1)^2}$$ , we see that $$\frac{1}{(p-1)^2} \equiv 1+2p+3p^2+4p^3 \bmod p^4$$ and we are done since it is easy to see by expansion that $$(2p-1)(2p-2)(2p-3)(1+2p+3p^2+4p^3) \equiv 2(-3+5p+p^2+p^3) \bmod p^4.$$

I does seem that for the OP function to bring an integer, the argument must be the smallest prime of a twin prime pair. But proving this seems very difficult. What about trying to disprove it by computer search of a counterexemple ?