The Expansion gives me: $6^{th}$ term = $\displaystyle\binom{n}{5} x^{n-5}y^5$ and $16^{th}$ term = $\displaystyle\binom{n}{15} x^{n-15}y^{15}$. What do I do next?
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$\dfrac{n!}{5!(n-5)!}=\dfrac{n!}{15!(n-15)!} $. So $n$ is $20$.
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For completeness $\left(\text{Courtesy:} \: \textbf{Did} ,\: \textbf{gandalf61} ,\: \textbf{bof} ,\: \textbf{Francis}\right)$ $\\\text{(*I agree this explanation still has a lot of details missing)}$
The coefficient of $x^ky^{n−k}$ is equal to the coefficient of $x^{n−k}y^k$. So we have $n−5=15n−5=15$ and hence $ n=20$.
$\underline{\text{Other solutions to}\:\: \displaystyle\binom{n}{5}=\binom{n}{15}\quad(**)}$
The binomial coefficient is defined as $$\displaystyle \binom{n}{k}=\begin{cases}\dfrac{n!}{k!(n-k)!} &0\leq k \leq n \\ 0 &\text{otherwise} \end{cases}$$ So by definition, $n=0,1,2,3,4$ are also solutions to $(**)$. $$\displaystyle \binom{n}{5}=\binom{n}{15}\implies \displaystyle \dfrac{15!}{5!}=\dfrac{(n-5)!}{(n-15)!}$$ If $z\in\mathbb{C}$ and $a,b\in\mathbb{Z}\:$ we have $$\dfrac{\Gamma(z-a+1)}{ \Gamma(z-b+1)}=(-1)^{b-a}\frac{\Gamma(b-z)}{ \Gamma(a-z)}$$ Here $a=5, b=15$. $$\dfrac{\Gamma(z-5+1)}{ \Gamma(z-15+1)}=(-1)^{15-5}\frac{\Gamma(15-z)}{ \Gamma(5-z)}=\prod_{i=5}^{14}(i-z)$$ So $\displaystyle \dfrac{15!}{5!}= \prod_{i=5}^{14}(i-z)$ has $10$ roots by fundamental theorem of algebra which are complex (requires more details).
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So for our question here it is reasonable to assume $n$ to be positive, $n\geq k$ and the expansion of $(x+y)^n$ to have $n+1$ terms.
So $\fbox{$n$=$20$}$
Yadati Kiran
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Yadati, how did you arrive at n=20.? Am a bit behind please. Usually, I have difficulties dealing with factorials of this nature – Bravie Nov 26 '18 at 09:41
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1@bof: True, but then you need to expand the binomial coefficient to complex plane through Gamma function, and it can get quite complicated I think. – Francis Nov 26 '18 at 09:46
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The coefficients in the binomial expansion are symmetric - the coefficient of $x^ky^{n-k}$ is equal to the coefficient of $x^{n-k}y^k$. So in this case we must have $n-5=15$ and hence $n=20$. – gandalf61 Nov 26 '18 at 09:52
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@gandalf61 You are repeating the argument showing that $n=20$ is a solution, but not advancing one iota in the direction of a proof that it is the only one. As a matter of fact, $n=20$ is not the only solution since $n=0$, $1$, $2$, $3$ and $4$ also are, in contradiction to your argument. – Did Nov 26 '18 at 10:00
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@Bravie: Its just by observation. There are 9 terms in between the $6^{th}$ and the $16^{th}$ term. The $10^{th}$ term seperates the expansion into two having equal number of terms. So we see there are $21$ terms in total. The number of terms in the expansion of $(x+y)^n$ is $n+1$. So $n$ is $20$. – Yadati Kiran Nov 26 '18 at 10:05
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@Did: I cannot agree with you more that $20$ is not the only solution. But are the other solutions relevant here? – Yadati Kiran Nov 26 '18 at 10:08
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@Did What about $n=-1$, is that also a solution, since $\binom{-1}k=(-1)^k$? – bof Nov 26 '18 at 10:11
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For an (incomplete) attack on the problem of the missing zeroes of the polynomial, see comment on main. To summarize, these zeroes exist all right, of course, but they seem to be either complex non real, or negative real. – Did Nov 26 '18 at 10:27
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@Did To make your additional solutions fit the original question you would have to argue that, for example, the expansion of $(x+y)^2$ has not just three terms but an infinite number of terms, and the coefficients of the sixth term in $x^{-3}y^5$ and the sixteenth term in $x^{-13}y^{15}$ are equal because they are both zero. That seems a very forced interpretation of the original question. I think it is reasonable to assume that $n$ is meant to be a positive integer here and the expansion of $(x+y)^n$ has just $n+1$ terms. – gandalf61 Nov 26 '18 at 10:52
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@gandalf61 You are missing the global picture here: to mention that $n=20$ solves this by symmetry of the binomial coefficients is true and incomplete. So we need another argument to show that there are no other solutions (together with the solutions $n=0$, $n=1$, $n=2$, $n=3$ and $n=4$, which the OP missed). Such an argument, which I presented in a comment (now inexplicably deleted) on main, uses the gamma function and proves that there are some other solutions to the relevant polynomial, but that these are complex not real, or real negative, ... – Did Nov 26 '18 at 11:02
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... hence, indeed, that they are irrelevant to the present context. (Amusingly, there exists another argument, somewhat simpler than the one I explained, but nobody mentioned it here yet.) Whichever way one chooses, considering what is actually posted on this page, it is a mathematical fact that one must add something to the pot before considering that the problem is solved. – Did Nov 26 '18 at 11:03
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"which are complex or real but negative" How do you know? FYI, this is the crucial step of this whole approach hence, if one cannot show this, all the rest falls. – Did Nov 26 '18 at 13:16
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I did find $z=20,-1$ is a solution. I am unable to compute or rather show that the remaining roots are complex. So I shall withdraw my answer. – Yadati Kiran Nov 26 '18 at 14:51
then n=0, n=1, n=2, n=3 and n=4 are solutions, taken from my comment above, possibly mean in your opinion? Sorry but posting random comments will not change the fact that your post below is not a complete answer. – Did Nov 26 '18 at 10:46