Assuming that $P(x)$ is a polynomial and it is true that $P(3x)=26 \cdot P(x+1)$, then of what degree is the polynomial?

Question

Assuming that $P(x)$ is a polynomial and it is true that $P(3x)=26 \cdot P(x+1)$, then of what degree is the polynomial?

This is the first time that I've seen such a question and hence I don't know how to attack it. What I attempted to do was the following:

If it is of degree 1, then $P(x)=ax+b$ hence $P(3x)=3ax+b$, but $P(3x)=26*P(x+1)$, hence \begin{align*} 3ax+b &= 26 \cdot 3a(x+1)+b \\ 3ax &= 26 \cdot 3a(x+1) \\ x & = 26 \cdot (x+1) \\ x &= 26 \cdot x+26, \end{align*}

Hence $x$ needs to be a certain value, something which can't be true.

I then tried the same thing for degree $2, 3, 4$ in the hope of finding some sort of pattern, but I couldn't find one. Could you please explain to me how to solve this question as well as how to solve other similar style questions?

Answer 1

Hint: $P(3x)=26 P(x+1)$ implies $3^n a_n = 26 a_n$ and so $a_n=0$ because $26$ is not an integer power of $3$.

Answer 2

When $P(x)\not\equiv0$ then there is an $n\geq0$ with $$P(x)=a_n x^n+\ldots+a_1 x+a_0,\qquad a_n\ne0\ .$$ It follows that $$Q(x):=P(x+1)=a_n x^n+b_{n-1}x^{n-1}+\ldots+ b_1 x+b_0$$ for certain $b_k$ determined by the $a_j$. When $x\ne0$ we can divide $$P(3x)\equiv26 P(x+1)=26 Q(x)$$ by $x^n$ and obtain $$3^n a_n +{3^{n-1}a_{n-1}\over x}+\ldots+{3a_1\over x^{n-1}}+{a_0\over x^n}=26\left(a_n+{b_{n-1}\over x}+\ldots+{b_0\over x^n}\right)\ .$$ This is valid for all $x\ne0$, in particular for $x\to\infty$. It follows that necessarily $(3^n-26)a_n=0$, hence $a_n=0$, contrary to assumption. This implies that in fact $P(x)\equiv0$. Therefore the degree of $P$ is $-\infty$.