# Combinatoric equation involving three unknowns

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?

• – D.W.
Commented Jun 5 at 20:42

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.