I want to prove
${2a \choose 2b} \equiv {a \choose b} \pmod 2$
with induction. I've seen one that uses the Iverson bracket notation, however that doesn't seem to fit into the scope of what I am learning right now. Is there another way to prove this?
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1Not that it matters if you don't want to use it, but do you mean the Iverson bracket? – Misha Lavrov Sep 05 '21 at 18:01
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You could try Lucas' Theorem. – lulu Sep 05 '21 at 18:03
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Yeah sorry I misspelled. – Yanbo hehe xd Sep 05 '21 at 18:08
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You can prove more generally that if $p$ is prime then $$\binom{pa}{pb}\equiv \binom{a}b\pmod{p^2}$$ But even more surprising is, the congruence is actually true $\pmod{p^3},$ when $p\geq 5.$ – Thomas Andrews Sep 05 '21 at 21:30
2 Answers
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You have $2a$ limes and you want to choose $2b$ of them. Divide your set into two buckets with $a$ limes in them a piece. A choice of $2b$ limes is called split if there are exactly $b$ limes in each bucket. Then ${a \choose b}^2$ gives the number of split choices. On the other hand, the number of non-split choices is even, because there is an involution of that set with no fixed-points given by flip-flopping the buckets.
So $$ {2a \choose 2b} = {a \choose b}^2 + \text{ something even }, $$ which proves the result since ${a \choose b}^2$ has the same parity as ${a \choose b}$.
hunter
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Use Pascal's identity
$$\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k} \, .$$
Apply this to $\displaystyle\binom{2a}{2b}$. Then apply it again to the two new binomial coefficients you just got. Then use the induction hypothesis.
Micah
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