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Let $p$ be a prime number and $s$ be a positive integer. Show that for any $i \in \{0, 1, . . . , p^s− 1\},\ \binom{p^s−1}i \equiv (−1)^i(mod\ p)$.

Formula-wise the question attempts to take the ratio of: $\frac{(p^s−1)!}{i!(p^s−1-i)!}$, for all $i\in \{0, 1, . . . , p^s− 1\}$.

On taking some special values of $i$, get the below:
1. If $i=0$, the ratio =$1$,
2. If $i=1$, the ratio =$p^s−1$,
3. If $i=p^s−1$, the ratio =$1$,

But am unable to understand what is the logical implication of the question; and why only prime value of $p$ is considered.

Also, suggest an approach to solve the same.

jiten
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    Proof by induction should work - look at the ratio between ${p^s-1}\choose i$ and ${p^s-1}\choose {i+1}$ and consider cases based on whether $i+1$ is a multiple of $p$. – Karl Mar 25 '20 at 05:12
  • @Karl But, why only prime values for $p$ is unclear. Also, any logical implication of the question. – jiten Mar 25 '20 at 05:15
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    The proof I have in mind uses the primality of $p$ to ensure that all the factors in the denominator that are not 0 mod $p$ have inverses mod $p$. What do you mean by logical implication? – Karl Mar 25 '20 at 05:37
  • @Karl By logical implication, mean that the property highlighted by the problem. I hope your comment shows the logical implication of the problem. Also, please post answer. – jiten Mar 25 '20 at 06:09

3 Answers3

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It suffices to show that $\binom{p^s-1}{i+1}\equiv(-1)\binom{p^s-1}i\pmod p$.

Write $i+1=p^kr$ where $p\not\mid r$. Since $p$ is prime and $p\not\mid r$, we have $\gcd(p,r)=1$, so there exists an integer $r^{-1}$ such that $r^{-1}r\equiv1\pmod p$. Also note that $k<s$, so $p^{s-k}\equiv0\pmod p$.

Without reducing mod $p$, we can verify the following equality of integers: $$ r\binom{p^s-1}{i+1}=(p^{s-k}-r)\binom{p^s-1}i $$

Then multiplying both sides by $r^{-1}$ and reducing mod $p$ using the observations above gives the desired result.

Karl
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  • Please elaborate how it is always possible to write $i+1 = p^kr$, where $p\not\mid r$. – jiten Mar 25 '20 at 07:19
  • Starting with $i+1$, keep dividing by $p$ until you can't. – Karl Mar 25 '20 at 07:19
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    In other words, take the largest $k$ such that $p^k\mid{i+1}$. – Karl Mar 25 '20 at 07:22
  • I hope you mean $$r\binom{p^s-1}{i+1}=(p^{s-k}-r)\binom{p^s-1}i=(-1)^{(i+1)}(mod \ p)$$ If am correct, still request elaboration of the same. – jiten Mar 25 '20 at 07:36
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    By "multiply both sides by $r^{-1}$ and reduce mod $p$", I mean you could write $\binom{p^s-1}{i+1}\equiv r^{-1}r\binom{p^s-1}{i+1}=r^{-1}(p^{s-k}-r)\binom{p^s-1}i\equiv r^{-1}(0-r)\binom{p^s-1}i\equiv(-1)\binom{p^s-1}i$. Note the difference between $=$ and $\equiv$. By "It suffices to show that X", I meant that once we've shown X, the requested result follows by induction. – Karl Mar 25 '20 at 19:00
  • Thanks for pointing out that the symbol $\equiv$ is used here for showing equivalent results (terms), rather than equality. It shows that induction is used to get equivalent terms. I hope then your last comment must be read as: $\binom{p^s-1}{i+1} \equiv r^{-1} r \binom{p^s-1}{i+1} \equiv r^{-1} (p^{s-k}-r)\binom{p^s-1}{i} \equiv r^{-1}(0- r) \binom{p^s-1}{i} \equiv (-1) \binom{p^s-1}{i}$. – jiten Mar 26 '20 at 00:00
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    The places where I used $=$ (including the one in my last comment) are actual equalities, meaning that two expressions denote the same integer. Equality implies modular congruence, so what you wrote is also true. Keep in mind that "$x\equiv y\pmod p$" is short for "$p\mid{x-y}$". – Karl Mar 26 '20 at 01:11
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I completely agree with Siong Thye Goh's answer and consider his answer more elegant than the one I am about to present. My answer is intended to show the much more pedestrian and inelegant but still viable approach of induction.

Assume that the conjecture is true for $i$, where $0 \leq i \leq p^s - 2.$
Then $\;\binom{p^s - 1}{i + 1},\;$ which is an integer, equals $\;\binom{p^s - 1}{i} \times \frac{p^s - [i+1]}{i + 1}.$

By inductive assumption, $\;\binom{p^s - 1}{i} \;\equiv (-1)^i \pmod{p}.$
Further, $\;(-1)^i [-(i+1)] \equiv (i+1)(-1)^{(i+1)} \pmod{p} \;\Rightarrow$
$\;\binom{p^s - 1}{i} \times [p^s - (i+1)] \;\equiv\; (i + 1) (-1)^{(i+1)}\pmod{p} \;\Rightarrow$
$\;\binom{p^s - 1}{i} \times \frac{p^s - [i+1]}{i + 1} \;\equiv\; (-1)^{(i+1)}\pmod{p}.$

$\underline{\text{Addendum}}$
As Karl indicated in his comment below, I must examine the case of $\;(i+1) \;\equiv\; 0 \pmod{p}\;$ separately. I am working on it and will update my answer if and when I find a remedy.

Convoluted remedy found : proof by contradiction.

Let $pk$ denote the smallest multiple of $p$ such that the conjecture is false.
Then, from the induction used in the first part of this answer, the conjecture is true for $i = (pk - 1).$

Let $v_p(n) \;\equiv\;$ the largest exponent $\theta$ such that $p^{\theta} | n.$
Let $v_p[(p^s - 1)!] = \alpha, \;v_p[(pk - 1)!] = \beta, \;v_p[(p^s - pk)!] = \gamma,\;$ and $\;v_p(pk) = \delta.$
By assumption, $\;\binom{p^s - 1}{pk-1} \;\equiv\; (-1)^{(pk-1)} \;\pmod{p}.$

Therefore, $\;\alpha = \beta + \gamma,\; (p^s - 1)!\;$ has form $\;(p^\alpha) \times a,$
$(pk-1)!(p^s - pk)!\;$ has form $\;(p^\alpha) \times b,$
where $p$ does not divide $a$ or $b$ and $a \equiv b\times (-1)^{(pk-1)} \pmod{p}.$

$v_p(pk) = \delta \;\Rightarrow v_p(p^s - pk) = \delta.$
Further, $\;pk = p^\delta \times r,\;$ where $p$ does not divide $r \;\Rightarrow\;$
$(p^s - pk)\;$ will have form $(p^\delta) \times (-t),\;$ where $\;t \;\equiv\; r \pmod{p}.$

Thus, $\;(pk)!(p^s - 1 - pk)! \;=\; $ $(pk) \times [(pk-1)!]\times\frac{(p^s - pk)!}{(p^s - pk)},$
which can be re-expressed as $(p^\delta \times r) \times (p^\alpha \times b)$ $\frac{1}{p^\delta\times(-t)} \;\Rightarrow\;$
$\binom{p^s - 1}{pk} \;\equiv\; (-1)^{(pk-1)} \times \frac{-t}{r} \;\equiv\; (-1)^{(pk)} \pmod{p}.$
This yields a contradiction. Therefore, the conjecture also holds for any $i < p^s - 1,$
where $i$ is a multiple of $p$.

user2661923
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  • Siong Thye Goh's answer was: \begin{align} \binom{p^s-1}{i} &= \frac{\prod_{j=0}^{i-1}(p^s-1-j)}{i!}\&\equiv \frac{\prod_{j=0}^{i-1}(-1-j)}{i!} \pmod{p}\ &\equiv \frac{(-1)^i i!}{i!} \pmod{p}\ &\equiv (-1)^i \pmod{p} \end{align}

    I don't think $p$ being prime is needed.

     – jiten Mar 25 '20 at 05:45
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    You're dividing both sides by 0 in the last step when $i+1$ is a multiple of $p$. The deleted answer was wrong for the same reason. – Karl Mar 25 '20 at 05:46
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    Notice that $\binom{6^1-1}2=\binom52=10\equiv4\not\equiv(-1)^2\pmod6$, so primality must be used. – Karl Mar 25 '20 at 05:50
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    Thanks for the counterexample. :) – Siong Thye Goh Mar 25 '20 at 05:54
  • @Karl Nice catch; see my addendum. – user2661923 Mar 25 '20 at 05:59
  • @Karl hopefully my remedy (in the addendum) is valid. – user2661923 Mar 25 '20 at 07:52
  • Please elaborate the significance of introducing four symbols in the line: Let $v_p[(p^s - 1)!] = \alpha, ;v_p[(pk - 1)!] = \beta, ;v_p[(p^s - pk)!] = \gamma,;$ and $;v_p(pk) = \delta.$ – jiten Mar 25 '20 at 08:12
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    @jiten $v_p(pk) = \delta;$ signifies that $pk$ can be expressed as $p^\delta \times r$, where $p$ does not divide $r$. The rest of the line signifies that $;\binom{p^s - 1}{pk - 1}$ can be expressed as $\frac{p^\alpha \times a}{p^{(\beta + \gamma)} \times b},;$ where $p$ does not divide $a$ or $b$. – user2661923 Mar 25 '20 at 08:17
  • I present my version of your comment: $v_p(pk)=\delta$ signifies that $pk$ can be expressed as $p^{\delta} \times r$, where $p\not \mid r$. Need the value of $\binom{p^s-1}{pk-1}$, for which in turn need $3$ terms (total of $4$): $p^s-1, pk-1, p^s-pk$. But, have a multiplier term of $v_p=\theta$ with these $3(4)$ terms, as $p^{\theta} \mid n$. Here, have an addtl. difficulty:: why $v_p$ is shown as multiplier to the four terms? – jiten Mar 25 '20 at 08:34
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    @jiten In your last comment, your 1st sentence expresses what I am trying to say. Unsure about your question in remainder of comment. I chose this approach, because in order to establish the contradiction expressed in my answer's addendum, I first have to determine (with respect to $\binom{p^s - 1}{pk - 1}$), the largest exponent $\theta_1$ such that $p^{\theta_1}$ divides the numerator. I similarly must find the largest exponent $\theta_2$ such that $p^{\theta_2}$ divides the denominator. Then, I have to establish the relationship between $\theta_1$ and $\theta_2$. See next comment. – user2661923 Mar 25 '20 at 08:58
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    @jiten If you examine the first part of my answer, and then examine Karl's subsequent comment (re my answer had a flaw), I must take special steps in the addendum to establish a contradiction. When completing the addendum, I was unaware of Karl's answer. If you compare my addendum to his answer, you will see very similar analysis. – user2661923 Mar 25 '20 at 09:00
  • I request your attention to the source of the problem, and the solution provided there: https://i.sstatic.net/6h1Nl.png. I am not sure how the author made the answer, but was not aware of the question's significance; so placed this question. – jiten Mar 25 '20 at 09:57
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    @jiten what you are referring to seems to be an elegant approach. Please re-read the first part of my answer. The whole point was to show that the inelegant induction approach, which (as in this case) often requires much less creativity is often viable. – user2661923 Mar 25 '20 at 10:07
  • I am still trying to understand your proof, and request to be in chat at : https://chat.stackexchange.com/rooms/105935/discussion-between-siong-thye-goh-and-jiten Will be waiting there for you to arrive. – jiten Mar 26 '20 at 06:23
  • I am in chatroom from here to many hours. – jiten Mar 27 '20 at 19:10
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Let $$v=\binom{p^s-1}{i} = \frac{\prod_{j=1}^{i}(p^s-j)}{i!}$$

Suppose $j \in \{1, i\}$, we can write $j=p^{r_j}w_j$ where $gcd(p,w_j)=1$, $r_j < s$, then $$p^s-j=p^s-p^{r_j}w_j=p^{r_j}(p^{s-r_j}-w_j)$$

$$v=\prod_{j=1}^i\left(\frac{p^s-j}{j}\right)=\prod_{j=1}^i\left( \frac{p^{r_j}(p^{s-r_j}-w_j)}{p^{r_j}w_j}\right)=\prod_{j=1}^i\left( \frac{p^{s-r_j}-w_j}{w_j}\right)$$

$$v\cdot \prod_{j=1}^iw_j = \prod_{j=1}^i (p^{s-r_j}-w_j)$$

$$v\cdot \prod_{j=1}^iw_j \equiv \prod_{j=1}^i (p^{s-r_j}-w_j) \equiv \prod_{j=1}^i (-w_j)\equiv (-1)^i \prod_{j=1}^iw_j \pmod{p}$$

Since $gcd(p,w_j)=1$, $\left(\prod_{j=1}^iw_j\right)^{-1}$ exists, multiplying it on the both sides, we have

$$v \equiv (-1)^i \pmod{p}$$

That is $$\binom{p^s-1}{i}\equiv (-1)^i \pmod{p}$$

Remark: I made a mistake earlier for not checking if an element is invertible.

This is the wrong approach:

\begin{align} \binom{p^s-1}{i} &= \left(\prod_{j=0}^{i-1}(p^s-1-j)\right)(i!)^{-1}\\ &\equiv \left(\prod_{j=0}^{i-1}(-1-j)\right)(i!)^{-1} \equiv (-1)^i(i!)(i!)^{-1} \\&\equiv (-1)^i \pmod{p} \end{align}

because I did not check that $(i!)^{-1}$ exists and indeed it need not exists.

In my latest approach, I ensure that $gcd(w_j,p)=1$ and hence $(\prod_j w_j)^{-1}$ exists. Also, by expressing $j=p^{r_j}w_j$, I have illustated that the $p^{r_j}$ factor in $p^s-j$ can cancel out with $p^{r_j}$ factor in $j$.

That is now I use the property that the

  • $p^{r_1}$ in $1$ and $p^s-1$ cancels out.
  • $p^{r_2}$ in $2$ and $p^s-2$ cancels out

and so on.

Siong Thye Goh
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  • How to show that the condition of $p$ being a prime is unneeded. – jiten Mar 25 '20 at 05:17
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    Does my proof use any prime property? – Siong Thye Goh Mar 25 '20 at 05:18
  • Please elaborate your first answer line, that shows it as a product of terms in the numerator. – jiten Mar 25 '20 at 05:19
  • I hope that once the condition for $p$ being a prime is shown as redundant, there is no special logic attached to the question; just an application of the binomial theorem. – jiten Mar 25 '20 at 05:20
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    The numerator is just $\frac{(p^s-1)!}{(p^s-1-i)!}$. Perhaps the intended proof uses stuff that involves prime number property like Wilson's theorem? Anyway, do check numerically for a few numbers. I haven't gave it sufficient thought yet and I might make mistakes. – Siong Thye Goh Mar 25 '20 at 05:23
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    I see, @jiten kindly remove the acceptance and I will delete the post. – Siong Thye Goh Mar 25 '20 at 05:29
  • Our discussion is in chat room : https://chat.stackexchange.com/rooms/105935/discussion-between-siong-thye-goh-and-jiten – jiten Mar 26 '20 at 04:05
  • Please be back in chatroom. As it is an abstract area, so have no idea about coding in it. Even a simpler program can help to build better ones. – jiten Mar 27 '20 at 01:07
  • Please be back in chatroom. Have failed as tried your wrong (but, easier) version, answer is $0$. Only your help can make it work by implementing correct version. – jiten Mar 27 '20 at 16:11
  • @jiten it is now 11:17am pacific standard time. i will be back in the chat room at 2pm pacific standard time. – user2661923 Mar 27 '20 at 18:18
  • My first version of correct code: https://py3.codeskulptor.org/#user305_zg4lA8zMse_2.py; yours' benchmark is at: https://py3.codeskulptor.org/#user305_JiODDeBcQcRCYUL.py – jiten Mar 28 '20 at 04:03
  • Please see my last comment to selected answer to my today's post at:https://math.stackexchange.com/q/3600042/424260 It attempts an extension of the question to $n^{34}-b$, where $b$ is not expressible as a square. – jiten Mar 29 '20 at 13:08
  • out of respect of the other users, it's not appropriate for me to interfere now. – Siong Thye Goh Mar 31 '20 at 09:41
  • Please be in chatroom for my updated answer for Q.6. Also, posted at : https://math.stackexchange.com/q/3606900/424260. – jiten Apr 02 '20 at 19:11
  • Please be in chatroom. – jiten Apr 03 '20 at 01:52