I want to compute $$\sum_{k=0}^{(p-1)/2} \binom{2k}{k} \pmod{p}$$ for odd prime $p.$ After checking several examples I think this is either $\pm 1$ though I haven't made much progress otherwise. I've considered writing $\binom{2k}{k} = \sum_{i=0}^{k} \binom{k}{i}^2$ but this doesn't do much. I also think maybe $\frac{1}{\sqrt{1-4x}}$ expansion might be useful along with $\binom{2n}{n} = 0 \pmod{p}$ when $n < p < 2n.$
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I thing using (first prove it using the same ideas that work for Taylor) that $\binom{2n}{n}\equiv (-4)^n\binom{\frac{p-1}{2}}{n}$ should do the trick. Idea from: https://artofproblemsolving.com/community/q1h2555421p21867440 – asking questions Aug 08 '22 at 09:14
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Does this answer your question? binomial congruence $\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}\equiv 0~or (-2)\pmod p$ - found using an Approach0 search. Adding for $1$ for the $i=0$ term gives $\pm 1 \pmod{p}$. There's also the AoPS thread Number theory. – John Omielan Aug 08 '22 at 13:25
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1@JohnOmielan Yep, looks like the same problem. I've also found similar problems posted on AoPS several times. Certainly interesting how many different approaches there are. https://artofproblemsolving.com/community/c6h2705505 https://artofproblemsolving.com/community/q1h2756659p24094700 – Roy Aug 08 '22 at 17:07
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Full solution after hint in comments:
Claim: $\binom{2n}{n}\equiv (-4)^n\binom{\frac{p-1}{2}}{n} \mod p$ for $0\leq n\leq \frac{p-1}{2}$.
Proof: Easy manipulation.
Now: $$\sum_{n=0}^{\frac{p-1}{2}}\binom{2n}{n}\equiv \sum_{n=0}^{\frac{p-1}{2}} (-4)^n\binom{\frac{p-1}{2}}{n}=(1-4)^{\frac{p-1}{2}}=(-3)^\frac{p-1}{2}.$$
So the answer is that the sum will be $0$ if $p=3$, $1$ if $-3$ is a quadratic residue mod $p$, and $-1$ if it isn't. Using quadratic reciprocity will give you a condition on what $p$ is mod $3$ to see whether it's $1$ or $-1$, i.e since $(-3/p)=(p/3)$ we get that the sum is $-1$ if $p\equiv 2$ mod $3$, and $1$ if $p\equiv 1$ mod $3$.
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1I think one way to see the claim is indeed to expand $(1-4x)^{-1/2}$ by binomial expansion and compare coefficients with $\binom{2n}{n}$ – Roy Aug 08 '22 at 09:57
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That's the idea I would follow, just note that $\binom{\frac{p-1}{2}}{k}=\binom{-1/2}{k}$, so the intuition holds. Moreover, for $x=1$ we have the square root of $-3$, which is what we want to see in the end. I guess by reverse engineering from $(-3)^{(p-1)/2}$ one could get the other expression as well using the Taylor idea. I'll leave the details of how they are related to you. – asking questions Aug 08 '22 at 10:07
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I think one might even be able to find a more combinatorial solution by invoking the well known identity $\sum \binom{2n}{n} \binom{2(n-k)}{n-k} = 4^n,$ which of course is itself closely related to the identity here. – Roy Aug 08 '22 at 10:38