Number Theory: How to prove this statement (Lucas's theorem?)

Question

Assume $1\le k \le p-1$ and $p$ is a prime number.

Prove:

a) ${p \choose k} \equiv 0$ (mod p)

b) ${p-1 \choose k} \equiv (-1)^k $ (mod p)

Probably the statement can be proved using Lucas's theorem. But I'm not sure about the right way to use if so. So any hints how to prove the statement?

Use first & http://mathworld.wolfram.com/PascalsFormula.html for the second — lab bhattacharjee, May 05 '17 at 18:27
The proof of Lucas's theorem that I have in mind depends of item a) so I would be a bit hesitant to use it here. Of course, you can also do this direction as an exercise — Jyrki Lahtonen, May 05 '17 at 18:33

score 2 · Answer 1 · answered May 05 '17 at 18:30

For (a) the fundamental theorem of arithmetic shows that $p|p!$ and since $p\not | k!$ for $1\le k\le p-1$ $p|{p\choose k}$ for $k$ in that range.

For (b) Use

$${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$$

Then with $n=p$ we see that

$${p-1\choose k-1}\equiv -{p-1\choose k}\mod p.$$

Since ${p-1\choose 1}\equiv -1\mod p$ the rest comes from induction.

score 2 · Answer 2 · answered May 05 '17 at 18:32

Lucas's theorem seems like overkill here. You'd also have to be careful, because a) is commonly used in the proof of Lucas's Theorem, so you'd need to have a proof of that theorem that doesn't use a).

For the first, ${p\choose k}=\frac{p!}{(p-k)!k!}$ has a numerator divisible by $p$ and a denominator that isn't, so the ratio is divisible by $p$.

For the second, you know that ${p-1\choose k-1}+{p-1\choose k}={p\choose k}=0\pmod{p}$. Now just induct on $k$ and use the fact that ${p-1\choose 1}=-1\pmod{p}$

score 0 · Answer 3 · answered May 06 '17 at 01:25

0

$\binom{p-1}{k} = \frac{(p-1)(p-2)...(p-k)}{k!} \mod p$

All terms in numerator are multiples of $p$ except one which is $= (-1)^k k!$

and so $\binom{p-1}{k} \mod p = (-1)^k k!/k! = (-1)^k \mod p$

answered May 06 '17 at 01:25

sku

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Number Theory: How to prove this statement (Lucas's theorem?)

3 Answers3