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$${p^{e+1} \choose p\cdot k } \equiv {p^{e} \choose k } \mod p^{e+1} $$ $p$ is prime, $e$ and $k$ are non negative integers.

I am struggling with a proof of the above proposition, in the case where $p=2$.

This is trivially true for $k=0$, then I use induction on $k$.

Let $k \gt 0 $ and suppose that $ {p^{e+1} \choose {p\cdot (k-1)} } \equiv {p^{e} \choose k-1 } \mod p^{e+1} $ (induction hypothesis)

${p^{e+1} \choose p\cdot k }={p^{e+1} \choose p\cdot (k-1))} \frac{p^{e+1}-k\cdot p +p }{k\cdot p} \frac{p^{e+1}-k\cdot p +p-1 }{k\cdot p-1}\cdot\cdot\cdot\frac{p^{e+1}-k\cdot p +1 }{k\cdot p-( p-1)} $

$({k\ p-1}) \cdot\cdot ({k\ p-( p-1))}{p^{e+1} \choose p\ k }={p^{e+1} \choose p\ (k-1))}\frac{p^e-k+1}{k}(p^{e+1}-k\ p +p-1) \cdot\cdot(p^{e+1}-k\cdot p +1) $ $({k\ p-1}) \cdot\cdot ({k\ p-( p-1))}{p^{e+1} \choose p\ k }\equiv{p^{e+1} \choose p\ (k-1))}\frac{p^e-k+1}{k}(-k\ p +p-1) \cdot\cdot(-k\cdot p +1) \mod p^{e+1}$

$({k\ p-1}) \cdot\cdot ({k\ p-( p-1))}{p^{e+1} \choose p\ k }\equiv{{p^{e} \choose k-1 }}\frac{p^e-k+1}{k}(-k\ p +p-1) \cdot\cdot(-k\cdot p +1) \mod p^{e+1}$
by induction hypothesis, then $({k\ p-1}) \cdot\cdot ({k\ p-( p-1))}{p^{e+1} \choose p\ k }\equiv{{p^{e} \choose k }}(-1)^{p-1}(k\ p -p+1) \cdot\cdot(k\cdot p -1) \mod p^{e+1}$
then, since the factors $ ({k\ p-1}), \cdot\cdot, ({k\ p-( p-1))}$ are not divisible by $p$, they can be removed and here is what is obtained : $${p^{e+1} \choose p\cdot k } \equiv (-1)^{p-1} {p^{e} \choose k } \mod p^{e+1} $$

which is the desired result when $p$ is odd, but when $p=2$ that is:
$${2^{e+1} \choose 2\cdot k } \equiv - {2^{e} \choose k } \mod 2^{e+1} $$ or $${2^{e+1} \choose 2\cdot k }- {2^{e} \choose k } \equiv - 2\cdot {2^{e} \choose k } \mod 2^{e+1} $$

The greatest power of 2 which divides ${2^{e} \choose k } $ is $2^{e -f}$, where $2^{f}$ is the greatest power of $2$ which divides $k$. Then if $k$ is odd, that is still ok, but if $k$ is even, $f\gt 0$ and the greatest power of $2$ which divides $- 2\cdot {2^{e} \choose k }$ is $2^{e+1 -f}\lt 2^{e+1}$. This cannot be zero modulo $2^{e+1}$.

But this is against the numerical evidence ex $${2^{2+1} \choose 2\cdot 2 }- {2^{2} \choose 2 }= 64 \equiv 0 \mod 2^{2+1} $$

Where am I wrong?

René Gy
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    More generally, $\dbinom{qn}{rn} \equiv \dbinom{qn/p}{rn/p} \mod p^k$, where $p$ is a prime, $q$ and $n$ are integers, $r$ is a rational number, and $k$ is a nonnegative integer such that $p^k \mid n$. Here, binomial coefficients $\dbinom{a}{b}$ with $b \not\in \mathbb{N}$ are understood to be $0$. See (11.360) in http://arxiv.org/src/1409.8356v3/anc/HopfComb-v54-with-solutions.pdf for a proof (actually, two proofs). – darij grinberg Jan 01 '16 at 23:04
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    Be careful with dividing by $k$; it might have common divisors with $p^{e+1}$. (This is exactly the case in which you are getting troubles.) – darij grinberg Jan 01 '16 at 23:11
  • Oh I see my mistake. That little illicit division by $k$ in the congruence also spoils my "induction proof" :-( . Thank you for the reference to a real proof... would you put your comments as an answer, I would accept and close the question. – René Gy Jan 02 '16 at 09:27

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Be careful with dividing by $k$; it might have common divisors with $p^{e+1}$. (This is exactly the case in which you are getting troubles.)

The problem itself is a particular case of a more general problem, which claims that $\dbinom{qn}{rn} \equiv \dbinom{qn/p}{rn/p} \mod p^k$, where $p$ is a prime, $q$ and $n$ are integers, $r$ is a rational number, and $k$ is a nonnegative integer such that $p^k \mid n$. Here, binomial coefficients $\dbinom{a}{b}$ with $b \not\in \mathbb{N}$ are understood to be $0$. See (11.360) in http://arxiv.org/src/1409.8356v3/anc/HopfComb-v54-with-solutions.pdf for a proof (actually, two proofs).

darij grinberg
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