Let $P(x)=a_0+a_1x+a_2 x^2+a_3x^3+…….+a_nx^n$ and $P(1)=4$ and $P(5)=136$

Question

Let $P(x)$ be a polynomial such that,
$$P(x)=a_0+a_1x+a_2 x^2+a_3x^3+…….+a_nx^n,~~(a_i,n\in{Z^{\geq 0}})$$

$$ P(1)=4, P(5)=136$$

We have to find $P(3)$

This problem is harder than it looks (at least for me)

What I tried to do was
$$P(1)=a_0+a_1+a_2+a_3+…….+a_n=4$$ and $$P(5)=a_0+5a_1+25a_2 +125a_3+…….+a_n5^n$$

Let $P(1)=S$, and we take $a_0$ to the side of $S$ and multiply $(S-a_0)$ by $5$ and some cancellations. Simply it leads nowhere

Can I get some Hints on how to proceed?

Answer 1

The crucial observation comes from the fact the coefficients need to be in $\mathbb{Z}^{\geq 0}$.

$P(5) = 136$ can only be written in the following ways using powers of 5:

• $1 + (27)(5)$
• $1 + (22-5i)(5) + (i+1)(5^2)$ for $i = 0,1,2,3,4$
• $1 + (2)(5) + (1)(5^3)$

The only one that satisfies $P(1) = 4$ is the last one which is $P(x) = 1 + 2x + x^3$.

Therefore, $P(3) = 34$