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Find $n\in\mathbb{N^*}$ and $p, q \in\mathbb{Q}$, such that: $$(pn+q)\binom{nk+1}{k}+(qn+p)\binom{nk+2}{k-1}+(pn+q)\binom{nk+1}{k-1}=\binom{nk+2}{k+1}$$

for $\forall k\in\mathbb{N^*}$

Here's my progress so far: $$(pn+q)(\binom{nk+1}{k}+\binom{nk+1}{k-1})+(qn+p)\binom{nk+2}{k-1}=\binom{nk+2}{k+1}$$ $$(pn+q)\binom{nk+2}{k}+(qn+p)\binom{nk+2}{k-1}=\binom{nk+2}{k+1}$$ $$(pn+q)\frac{(nk+2)!}{k!(nk+2-k)!}+(qn+p)\frac{(nk+2)!}{(k-1)!(nk+3-k)!}=\frac{(nk+2)!}{(k+1)!(nk-k+1)!}$$ $$\frac{pn+q}{k!(nk+2-k)!}+\frac{qn+p}{\frac{k!}{k}(nk+3-k)!}=\frac{1}{(k+1)k!(nk+1-k)!}$$ $$\frac{pn+q}{(nk+1-k)!(nk+2-k)}+\frac{k(qn+p)}{(nk+1-k)!(nk-k+2)(nk-k+3)}=\frac{1}{(k+1)(nk+1-k)!}$$ $$(k+1)(nk-k+3)(pn+q)+k(k+1)(qn+p)=(nk-k+3)(nk-k+2)$$

Now I am stuck. If this is true $\forall k\in\mathbb{N^*}$, should $n,p,q$ be given in terms of $k$ aswell? Or are there exact solutions for this problem?

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So after expanding the brackets, moving everything to one side etc., we get: $$k^2(pn^2-np+2nq+p-q-n^2+2n-1)+k(pn^2+2np+2nq+p+2q-5n+5)+(3np+3q-6)=0$$ If this is true $\forall k\in\mathbb{N^*}$, this means that $a=b=c=0$ given the equation $ak^2+bk+c=0$

$pn^2-np+2nq+p-q-n^2+2n-1=0$ (1.)

$pn^2 +2np+2nq+p+2q-5n+5=0$(2.)

$3np+3q-6=0$, or $np+q=2$

Substracting (1.) from (2.) results in:

$$-3np-3q-n^2-7n-6=0$$ $$n^2-7n+12=0$$ Which implies $n_1=3, n_2=4$

If $n=3$, then: (1.): $9p-3p+6q+p-q-9+6-1=0$, or $7p+5q=4$

And also $3p+q=2$

So $p_1=\frac{3}{4}, q_1=-\frac{1}{4}$

For $n=4$, we have: (1.)

$13p+7q=9$

$4p+q=2$

Which results in $p_2=\frac{1}{3}, q_2=\frac{2}{3}$

This implies that the condition $p,q\in\mathbb{Q}$ was just the result of the equation.

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