Find the degree of the polynomial p(x) which satisfies the condition (x)p(x – 1) = (x – 15) p(x). So here is a problem from quadratics section:
Find the degree of the polynomial p(x) which satisfies the condition
$$(x)p(x – 1) = (x – 15) p(x)$$
My approach
Attempt 1:
I first went for method of undertermined coefficients setting
$$p(x)=a_0x^n+a_1x^{n-1}+...+a_{n-1}x+a_n$$
$p(x)=\frac{x}{x-15}p(x-1)\quad$-$\quad$(i)
$\implies\quad p(0)=p(1)=0$
To be honest I couldn't frame any meaningful condition

Attempt 2:
Went on to formulate a recursive formula which is clearly hinted by the expresion but couldn't make out of it.
I simply looked and found it to be first order linear recurrence relation.
$$q(x)=\frac{x}{x-15}\quad and\quad f(x)=q(1)q(2)q(3)...q(x)$$
Dividing eqn (i) throughout by f(x)
$$\frac{p(x)}{f(x)}=\frac{q(x)p(x-1)}{f(x)}$$
$$\frac{p(x)}{f(x)}=\frac{p(x-1)}{f(x-1)}$$
$$a_n=\frac{p(n)}{f(n)}$$
$$a_r - a_{r-1} = 0 $$
Summing all from r=1 to r=n
$$a_n=a_1$$
But $p(1)=0$ so is $a_1=0$
$$\implies\quad p(x)=0$$
Guide the solution with thinking approach as this is not the complete solution.
Edit: Answer is $P(x) = ax(x – 1) (x – 2) ... (x – 14)$
 A: Write
$$\frac{p(x)}{p(x-1)} = \frac{x}{x-15}$$
Now try to express $\frac{x}{x-15}$ as $\frac{g(x)}{g(x-1)}$. With some guesswork, you conclude that
$g(x) = x(x-1)\cdots (x-14)$ works. So you have
$$\frac{p(x)}{p(x-1)} = \frac{g(x)}{g(x-1)}$$
From here you conclude
$$\frac{p(x)}{g(x)} = \frac{p(x-1)}{g(x-1)}$$
that is, the rational fraction
$$r(x) \colon = \frac{p(x)}{g(x)}$$
satisfies $r(x) = r(x-1)$. From this we conclude that $r(x) = c$ constant. Conclusion
$$p(x) = c \cdot x(x-1) \cdots (x-14)$$
A: Clearly if $p(x)$ is a constant, $p(x)=0$ and this is a valid solution.  Else $p(x)$ must have a finite degree $n$, and as many number of roots.
Let $\alpha$ be a root of $p(x)$. Then as $\alpha \,p(\alpha-1) = 0$, we must either have $\alpha=0$ or $\alpha-1$ also to be a root.  In the latter case, recursively there would be more roots, and as the number of roots must be finite, this can terminate only with the case $\alpha=0$.  Hence the roots must be of form $0, 1, 2, \cdots n-1$.
Further putting $x = \alpha+1$, we get $0 = (\alpha-14)\, p(\alpha+1)$, which by a similar logic means unless $\alpha=14$, we must have $\alpha+1$ also as a root.  Hence $n-1=14$, so the degree is $15$, considering the least multiplicity for all roots.
Further we can show that the multiplicity of the roots cannot exceed $1$, consider $p(x-1)+xp'(x-1) = p(x)+(x-15)p'(x)$.  Suppose $\alpha$ is a multiple root, i.e. $p(\alpha)=p'(\alpha)=0$, then we have $p(\alpha-1)+\alpha p'(\alpha-1)=0$.  Now $\alpha=0$ is clearly not possible, hence $\alpha-1$ is also a multiple root, and this doesn't terminate, so $p'$ does not have a finite number of roots, which is a contradiction...
Hence the only non-zero polynomial is in fact of form $p(x) = a\,x\,(x-1)(x-2)\cdots (x-14)$, having degree $15$.
A: Set $x=0$. You get $0=0\cdot p(-1)=-15p(0)$ so $p(0)=0$.
Now set $x=1$. You get $0=1\cdot p(0)=-14p(1)$ so $p(1)=0$.
Now, set $x=2$. You get $0=2\cdot p(1)=-13p(2)$ so $p(2)=0$.
And so on and so forth, until, in the last step, you set $x=14$ and get $0=14p(13)=(-1)p(14)$ so $p(14)=0$.
This means that $p(0)=p(1)=\cdots=p(14)=0$, and so $p$ must be divisible by $x(x-1)(x-2)\cdots (x-14)$, and so, if nonzero, it must be of degree at least $15$.
On the other hand, $p_0(x)=x(x-1)(x-2)\cdots (x-14)$ already satisfies this equality because $p_0(x-1)=(x-1)(x-2)\cdots(x-15)$ and indeed $xp_0(x-1)=(x-15)p_0(x)$.
So, $15$ is the smallest degree of the non-zero polynomial $p$ which satisfies the given equation.

Edited: One can actually see that every nonzero polynomial $p$ satisfying the above equation is actually of degree $15$ and is the same as $p_0$ above up to a constant factor.
Namely, we have already concluded that $p$ must be divisible by $p_0$ above, i.e. $p=p_0q$ where $q$ is another polynomial. What does it mean for our equation?
$$xp_0(x-1)q(x-1)=(x-15)p_0(x)q(x)$$
As $xp_0(x-1)=(x-15)p_0(x)$ already, we can cancel that factor on both sides and obtain:
$$q(x-1)=q(x)$$
which is only possible if $q$ is a constant polynomial $q=C\in\mathbb R$, i.e. $p$ is the same as $Cp_0=Cx(x-1)(x-2)\cdots(x-14)$ for some constant $C$.
