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Trying to get a simple solution for the following secondary school problem:

Numbers +1 and -1, total amount $2^k$, are arbitrary placed along a ring. Then between any two of neighbouring numbers their product is placed, while the original numbers are wiped out. This is repeated again and again. Prove that in no more then $2^n$ of such operations only 1-s will remain.

Here is my own solution, that looks too complex to me. Maybe someone can suggest smth simpler.

Use $\{a_i\}$ $(0\le i\lt 2^k)$ for the original numbers (ordered clockwise), and $\{b_i^{(m)}\}$ $(0\le i\lt 2^k)$ the numbers after $m$-th operation, while $b_i^{(0)}=a_i$ $(0\le i\lt 2^k)$

Also $\oplus$ will be used for 'cyclic edition', ie addition by modulo $2^k$ (in other words $a \oplus b$ is the remainder of $a+b$ divided by $2^k$: $a \oplus b = (a+b) \& (2^k-1)$). Therefore case $a_{i \oplus 1}$ is the clockwise neighbour of $a_i$, $a_{i \oplus 2}$ is the second clockwise neighbour of $a_i$, etc. In particular $a_{i \oplus 2^k} = a_i$ (full-ring distance).

Prove (by induction) that $$ b_i^{(m)} = \prod_{j=0}^{m} a_{i \oplus j}^\binom{m}{j} \qquad (1) $$

where $\binom{m}{j}$ are binomial coefficients.

Base For $m=0$, expression $(1)$ becomes (assuming $\binom{0}{0}=1$) $b_i^{(0)} = a_i$, which is true.

Transition If $(1)$ folds for $m$, then

$$ b_i^{({m+1})} = b_i^{(m)}b_{i\oplus 1}^{(m)} = \prod_{j=0}^{m} a_{i \oplus j}^\binom{m}{j} \times \prod_{j=0}^{m}a_{i \oplus j \oplus 1}^\binom{m}{j} = $$ $$ \prod_{j=0}^{m} a_{i \oplus j}^\binom{m}{j} \times \prod_{j=1}^{m+1}a_{i \oplus j}^\binom{m}{j-1} = a_i \times (\prod_{j=1}^{m} {a_{i \oplus j}^{\binom{m}{j}+\binom{m}{j-1}}}) \times a_{i \oplus m \oplus 1} $$

Finally, due to $\binom{m}{j}+\binom{m}{j-1}=\binom{m+1}{j}$ and $\binom{m+1}{0}=\binom{m+1}{m+1}=1$ we get $$ b_i^{({m+1})} = \prod_{j=0}^{m+1} a_{i \oplus j}^{\binom{m+1}{j}} $$ to complete that proof of (1).

Substituting $m=2^k$ gives $$ b_i^{({2^k})} = \prod_{j=0}^{2^k} a_{i \oplus j}^{\binom{2^k}{j}} = a_i \times ( \prod_{j=1}^{2^k-1} a_{i \oplus j}^{\binom{2^k}{j}} ) \times a_{i \oplus 2^k} $$

But $a_i = a_{i \oplus 2^k}$, therefore $$ b_i^{({2^k})} = a_i^2 \prod_{j=1}^{2^k-1} a_{i \oplus j}^{\binom{2^k}{j}} $$

Since all the binomial coefficients, except first and last for power $2^k$ are even (see this thread), all $b_i^{({2^k})}$ are ones, qed.

This is certainly not the solution meant for a secondary school student :)

cyanide
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2 Answers2

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It is easy to notice the pattern that you are following Pascal's triangle in the terms which accumulate in the product.

It is fairly easy to notice that for early powers of $2$ the entries in the line are all even apart from the end ones, and you can't control the sign of those - the squares arising from the even numbers will all be $+1$.

It is easy to observe that when you get to the $2^n$th term, the two ends reference the same original number reached in opposite directions (clockwise and counterclockwise), and therefore make a final square giving $+1$.

Then there is the question of proving it - the Pascal's triangle bit is easy. The fact that the entries in the relevant rows are all even may be a little more challenging at high-school level.

You can do the evenness of the binomial coefficients as follows with $p,q$ polynomials:

$$(x^n+2xyp(x,y)+y^n)^2=x^{2n}+4x^2y^2p(x,y)^2+y^{2n}+2\times \text{ cross term multiples of }xy=$$$$=x^{2n}+2xyq(x,y)+y^{2n}$$

Start with $x^2+2xy+y^2$ ie $p(x,y)=1$ and use induction.

Mark Bennet
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  • The coherence of Pascal Triangle becomes clear, when instead of numbers we use their (-1) powers, i.e. replace 1-s with 0-s and (-1)s with 1-s, using exclusive or (a xor b = (a+b) & 1) instead of product. Still, rather tricky to prove that we eventually get all 0-s – cyanide Nov 10 '17 at 12:48
  • Certainly the trickiest part is the evenness of non-edge binomial coefficients. – cyanide Nov 10 '17 at 12:50
  • @cyanide I have added an elementary way of doing the evenness by induction. I think it is possible once you have identified that this is what has to be proved. – Mark Bennet Nov 10 '17 at 14:28
  • Actually I had something similar in mind. (More awkward than yours, but effectively the same). However, I prefer a completely different and much simpler idea. I'll wait for couple of days, if nothing else comes, will mark it as an answer. – cyanide Nov 10 '17 at 19:27
  • @cyanide spurred on by your encouragement I've posted a different idea which works more naturally with a circle than with the closely related linear string. – Mark Bennet Nov 10 '17 at 19:57
  • BTW, I found a very simple proof that $\binom{p^n}{i}$ ($p$ prime) is divisible by $p$ if $0<i<p^n$. See https://math.stackexchange.com/questions/51469/prime-dividing-the-binomial-coefficients/2519411#2519411 – cyanide Nov 14 '17 at 05:46
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I think you can also do this by a different induction. After two steps the value at position $i$ is the product of the values at positions $i-1$ and $i+1$ (taking account of wrap-around, the value at $i$ gets squared).

This decouples the circle into two versions of the problem for $n-1$ provided you do two steps at a time instead of one. ie twice what we had for $n-1$.

The base case is easy.

Mark Bennet
  • 100,194