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Let $n$ be a positive integer. What is the largest integer $f(n)$ such that there exists $p\in \mathbb{Z}[x, y]$ such that $$(x+2y)^{2n}-x^{2n}=2^{f(n)} p?$$

I have computed it for some small values and my conjecture is that $f(n)$ is the order of divisibility of $n$ by $2$ plus $2$ (e.g. $f(1)=2$). Is this correct?

redob
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    It certainly can't be any larger than that, because the highest-order term in $x$ for this polynomial is $4n x^{2n-1} y$. – eyeballfrog Sep 07 '19 at 20:06
  • @eyeballfrog well noted. – redob Sep 07 '19 at 20:09
  • But divisibility of $\binom{2n}{k}$ by $2n$ is not assured for all $k$ (see for instance https://math.stackexchange.com/questions/545962/when-is-binomnk-divisible-by-n?noredirect=1&lq=1 ), so we could check whether $n$ divides this coefficient since the $2^k$ would take care about the rest. Yet I think this is still quite hard, but I don't know either how to compute easily $v_2(\binom{2n}{k})$ (where $v$ is Legendre's formula). – zwim Sep 07 '19 at 20:27
  • @zwim I think you mean "where $v_2$ is the $2$-adic valuation", or similar. Legendre's formula is a statement about the $2$-adic valuation of the factorial of a number. – diracdeltafunk Sep 07 '19 at 21:14

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By the binomial theorem, the left-hand side is simply

$$\sum_{i=1}^{2n} \binom{2n}{i} 2^i y^i x^{2n-i},$$

so $f(n)$ is just

$$\min_{1 \leq i \leq 2n} v_2\left(\binom{2n}{i} 2^i\right) = \min_{1 \leq i \leq 2n} \left(i + v_2 \binom{2n}{i}\right),$$

where $v_2(x)$ is the number of times $2$ divides $x$. So, as noted by another commenter, you have found an upper bound for $f(n)$ (simply choose $i = 1$ to get $i +v_2\binom{2n}{i} = v_2(n)+2$). By Kummer's Theorem, $v_2\binom{2n}{i}$ is the number of carries performed when adding $i$ to $2n-i$ in binary. We can use this to show that the upper bound you gave is indeed the exact value of $f(n)$.

First, suppose $i \geq 2^{v_2(2n)}$. Then $$i + v_2\binom{2n}{i} \geq 2^{v_2(2n)} = 2 \cdot 2^{v_2(n)} \geq 2(v_2(n)+1) = 2v_2(n) + 2 \geq v_2(n) + 2,$$ as desired.

Now consider the case where $1 \leq i < 2^{v_2(2n)}$. The binary expansion of $2n$ ends with exactly $v_2(n)+1$ zeros, and by assumption the binary expansion of $i$ has at most $v_2(n)+1$ digits. Thus, the last $v_2(n)+1$ binary digits of $2n-i$ are simply the binary digits of $2^{v_2(n)+1} - i$, and the number of carries performed when adding $i$ to $2n-i$ is at least the number of carries performed when adding $i$ to $2^{v_2(n)+1} - i$. By thinking through the addition algorithm, this is seen to be simply $v_2(n) - v_2(i) + 1$. So we want to show that $i + v_2(n) - v_2(i) + 1 \geq v_2(n) + 2$, which is true since $i > v_2(i)$ for all positive integers $i$.

diracdeltafunk
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