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Are there integers is $i,j \geq 1$ for which there is an integer $k \geq 0$, such that:

$$\frac{(i+j-1)!}{i!(j-1)!}=2^k ?$$

There are two trivial families of solutions: For $i=1$:

$$ \frac{j!}{(j-1)!}=j $$

and whenever $j$ is a power of 2 we have a solution $(1,j)$. And, for $j=2$:

$$ \frac{(i+1)!}{i!} = i+1$$

and whenever $i+1$ is a power of 2 we have the second family of solutions (i,2). Are there any other solutions? And if not, is there a proof?

EDIT: As was pointed out, one can write the equation as a binomial coefficient:


with $n=i+j-1$ and $m=i$. Since the product of $m$ consecutive integers greater than $m$ is divisible by a prime $p$ greater than $m$, we have that

$$\binom{n}{m}= \frac{n(n-1)\ldots (n-m+1)}{m!} $$

is divisible by a prime $p$ and can therefore not be a power of 2 (if $m,n-m\geq 2 \Leftrightarrow j,i-1 \geq 2$). The two families of solutions given above are therefore the only solutions to this equation.

See also: Binomial Coefficients that are powers of 2


marked as duplicate by lulu, Community Aug 7 '17 at 19:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Thanks for the hint! I think my question is indeed a duplicate of the one you linked. I will try to paraphrase the proof given there and then mark my question as answered :) $\endgroup$ – Konrad Tschernig Aug 7 '17 at 18:53


$$ \left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n!}{r!(n-r)!} \\n=i+j-1 ,r=j \to (n-r)=i+j-1-i=j-1\\\frac{(i+j-1)!}{i!(j-1)!}= \left(\begin{array}{c}i+j-1\\ i\end{array}\right)=\left(\begin{array}{c}i+j-1\\ j-1\end{array}\right)$$this mean the "how many solution for "$x_1+x_2+...+x_j=i$ where $x_i \in \mathbb{Z}^+$

or , you can look at pascal triangle ,and find $2^k$ forms \begin{array}{rccccccccc} n=0:& & & & & 1\\\noalign n=1:& & & & 1 & & 1\\\noalign n=2:& & & 1 & & 2 & & 1\\\noalign n=3:& & 1 & & 3 & & 3 & & 1\\\noalign n=4:& 1 & & 4 & & 6 & & 4 & & 1\\\noalign \end{array}

for example $$x_1+x_2=0 \to (0,0) $$is only solution $$\to 2^k=1 \to k=1\\i=0,j=2$$ or $$x_1+x_2=3 \to (3,0),(2,1),(1,2),(0,3) \to 2^k=4 \to k=2\\j=2,i=3$$ or $$x_1+x_2=7 \to 8-solution \space 2^k=8 \to k=3 \\j=2,i=7$$or $$x_1+x_2+...+x_8=1 \to 8-solution \space 2^k=8 \to k=3 \\j=8,i=1$$ and so on

  • $\begingroup$ Hey, thanks for answering so quickly! I think that the solutions you point out are already contained in the two "trivial cases" I mentioned above. I suspect that there are no other solutions, but is it possible to prove this? $\endgroup$ – Konrad Tschernig Aug 7 '17 at 18:38
  • $\begingroup$ I found the proof: (see edit in my post). Your hint for rewriting the expression in terms of a binomial coefficient was indeed the way to victory! :) $\endgroup$ – Konrad Tschernig Aug 7 '17 at 19:13

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