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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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $$k + m = 2m(k-m)$$

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  • $\begingroup$ 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

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