Find all positive integers $n,k$ such that $$\binom{n-m}{k+m}=\binom{n+m}{k-m}$$

1) I solved problem if $m=1$. Its here: $k=1; n=3$

2) $$\binom{n-m}{k+m}=\binom{n+m}{k-m}$$ $k=m, n=3m$ is root of this equation.

Does this equation have other roots?

  • 4
    $\begingroup$ Numerical search for binomial coefficients up to $10000000$ only found the non-integral solution $n=\frac{29}{2},\, k=\frac{11}{2},\,m=\frac{1}{2}$ $\endgroup$ – Peter Taylor Jul 13 '17 at 15:02
  • $\begingroup$ Are you asking for a fixed $m$ what are the solutions? $\endgroup$ – alex.jordan Jul 15 '17 at 7:27
  • $\begingroup$ @alex.jordan: Yes exactly. $m$ is fixed. $\endgroup$ – Roman83 Jul 15 '17 at 7:57
  • $\begingroup$ Are you interested only in n,m,k integers or also in n,m,k reals? you tagged it with integer so I would assume just integers right? $\endgroup$ – zen Jul 17 '17 at 11:22
  • 3
    $\begingroup$ The non-integer solution found by @PeterTaylor is part of a larger family described by: $m=\frac{1}{2}$, $k=F_{2i}F_{2i+3}+\frac{1}{2}$, $n=F_{2i}F_{2i+5}+\frac{3}{2}$, where $F_i$ denotes $i$-th Fibonacci number. $\endgroup$ – Peter Košinár Jul 22 '17 at 8:06

I can't figure out how to get the roots, but I have tried for an approximation. Starting with $$\binom{n-m}{k+m} = \binom{n+m}{k-m}$$ This is $$\frac{(n-m)!}{(k+m)!(n-k-2m)!} = \frac{(n+m)!}{(k-m)!(n-k+2m)!} =$$ $$\frac{(n-m)(n-m-1)...(n-k-2m+1)}{(k+m)!} = \frac{(n+m)(n+m-1)...(n-k+2m+1)}{(k-m)!}$$ Note that the first numerator has $k+m$ products while the second has $k-m$ products. $$Numerator_1(k-m)! = Numerator_2(k+m)!$$ $$Numerator_1 = Numerator_2(k+m)(k+m-1)...(k-m+1)$$ There are $2m$ products in the expression following numerator 2. Thus, for $k$ and $n$ large, $m$ and $k$ approximately satisfy $$k + m = 2m(k-m)$$

  • $\begingroup$ I understand your answer all of it except this part $\frac{(n-m)(n-m-1)...(n-k-2m+1)}{(k+m)!} = \frac{(n+m)(n+m-1)...(n-k+2m+1)}{(k-m)!}$, could you please explain to me what you did in this step ? $\endgroup$ – anas pcpro May 24 at 15:29
  • $\begingroup$ Since $(n-m)! > (n-k-2m)!$, I eliminated $(n-k-2m)!$ from the denominator. The last product before $(n-k-2m)$ is $(n-k-2m+1)$, then you have $(n-k-2m)!/(n-k-2m)!$. Similar approach for the second fraction. $\endgroup$ – Vahan May 24 at 16:44
  • $\begingroup$ Thanks for explanation. $\endgroup$ – anas pcpro May 25 at 14:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.