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Let $A:=\sum_{k=0}^{\infty}x^{2^k}$. For what $n$ is it true that

$(A+1)^n+A^n\equiv1\mod2$ (here we are basically working in $\mathbb{F}_2$.)

The answer is all powers of 2, and it's fairly simple to see why they work, but the hard part is proving all non-powers of 2 don't work. In the solution, it says

"If $n$ is not a power of 2, say $n=2^i(2j+1),j\ge1$, then the smallest m for which $\binom{n}{m}$ is odd is $2^j$."

I don't see why. I know the highest power of $2$ in $\binom{n}{m}$ is $\sum_{k\ge1}[n/2^k]-[m/2^k]-[(n-m)/2^k]$ where $[.]$ is the greatest integer function. Also each summand is 0 or 1. It's 1 when the sum of fractional parts $\{m/2^k\} + \{(n-m)/2^k\}\ge1$, and 0 otherwise.

user70520
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2 Answers2

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The parity of a binomial coefficient can be quickly checked using Lucas' theorem.

Let $p$ be a prime. Write $n$ and $k$ in base $p$ as $$ n=\sum_{\ell\ge0}n_\ell p^\ell,\qquad k=\sum_{\ell\ge0}k_\ell p^\ell $$ with $0\le n_\ell,k_\ell <p$ for all $\ell$. Then we have the congruence $$ {n\choose k}\equiv\prod_{\ell\ge0}{n_\ell\choose k_\ell}\pmod p, $$ where we interpret a binomial coefficient as zero, if $k_\ell>n_\ell$ for some $\ell$. For a proof see for example this answer.

In your case $p=2$, and the congruence tells you the parity: the binomial coefficient ${n\choose k}$ is odd, if and only if $n_\ell=1$ whenever $k_\ell=1$.

You have $n=2^i(2j+1)$, so we see that $n_\ell=0$ for all $\ell<i$ and $n_i=1$. Therefore for the binomial coefficient ${n\choose k}$ to be odd, we must have $k_\ell=0$ for all $\ell<i$ also. The smallest positive integer $k$ with this property is, of course $k=2^i$. As then $k_\ell=0$ for all $\ell>i$, Lucas' theorem then tells us that ${n\choose k}$ is, indeed, an odd integer.

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Jyrki Lahtonen
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  • An occasionally useful result supplementing Lucas theorem is: add the numbers $k$ and $n-k$ together in the grade school pencil & paper method - using base $p$. Then ${n\choose k}$ is divisible by $p$, iff there is at least one "carry digit" in the calculation. Actually the number of carries gives you the power of $p$ dividing the binomial coefficient. In the multinomial case (when you add several integers together) the sum of the carries gives the exponent of $p$. – Jyrki Lahtonen Aug 11 '13 at 13:06
  • @user70520: For a proof of the result in the preceding comment see this answer. It took me while to find one here, sorry about that. – Jyrki Lahtonen Aug 21 '13 at 18:40
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Slightly variant tact: observe that if $m>1$ is odd then in ${\bf F}_2[[x]]$ we may compute

$$\left(1+\sum_{k=0}^\infty x^{2^k}\right)^{2^lm}+\left(\sum_{k=0}^\infty x^{2^k}\right)^{2^lm}=\left(1+\sum_{k=l}^\infty x^{2^k}\right)^m+\left(\sum_{k=l}^\infty x^{2^k}\right)^m\equiv 1+x^{2^l}~\bmod x^{2^l+1}. $$

anon
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  • anon: Are you sure about your identity (the one with the = sign)? For $l=1,m=3$ it fails. – user70520 Aug 12 '13 at 01:37
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    @user70520 The equality holds in ${\bf F}_2[[x]]$ (not ${\bf Z}[[x]]$). It follows from (an inductive version of) "freshman's dream" in characteristic two: $$\left(\sum_k a_k\right)^{2^l}=\sum_k a_k^{2^l}.$$ – anon Aug 12 '13 at 03:09
  • Thanks anon. How would you show it's true when each k belongs to a countable set (like $\mathbb{N}$)? Induction only proves when the set is finite. – user70520 Aug 13 '13 at 02:45
  • @user70520 I was speaking of induction on $l$. One only needs to prove freshman's dream for infinite sums to get the ball rolling: $$\left(\sum_k a_k\right)^2=\sum_{j,k}a_ja_k=2\left(\sum_{i<j}a_ia_j\right)+\sum_ka_k^2=\sum_k a_k^2.$$ – anon Aug 13 '13 at 05:48
  • Yeah that works too. Thanks. Just wondering how would you prove the infinite case from the finite version $(\sum_{k=1}^na_k)^2=\sum_{k=1}^na_k^2+2\sum_{1\le i<j\le n}a_ia_j$ (i.e., $n\to\infty$)? – user70520 Aug 13 '13 at 07:08
  • @user70520 I see no reason to prove it from the finite version. If you wanted to do that you'd have to use topology (e.g. the $(x)$-adic topology on ${\bf F}_2[[x]]$). – anon Aug 13 '13 at 07:11
  • Then how to prove the identity for infinite sum? Thank you. – user70520 Aug 13 '13 at 07:30
  • @user70520 I am not sure what you mean; I just proved the identity for infinite sums two comments ago. Could you elaborate? Are you asking, as a follow-up to my latest comment, how to prove the infinite version from the finite version using the $x$-adic topology, or what? – anon Aug 13 '13 at 07:38
  • How would you show $(\sum_ka_k)^2=\sum_ka_k^2+2\sum_{i<j}a_ia_j$ where the sum is not finite? Thanks. – user70520 Aug 18 '13 at 01:54
  • @user70520 By multiplying it out and rearranging terms, as I did above. If you understand the manipulations for a finite number of terms then you understand the manipulations for an infinite number of terms; barring convergence issues (which aren't worrisome here) they are just as valid, since they apply to the partial sums. – anon Aug 18 '13 at 01:55