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Let $P(x)$ be a third-degree polynomial with coefficients as natural numbers, and the constant term of $P(x)$ is $1$, and the sum of the coefficients of $P(x)$ is $2020$. Prove that there exist positive integers $a$ such that $(P(a))^2$ and $P(a)$ have different units digits when written in the decimal system.

This question comes from the book: Đa thức (Polynomial) - Nguyen Tien Lam.

Let P(x) = $mx^3+nx^2+px+1$ and $m+n+p=2019$, and i can't go any further.

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2 Answers 2

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The problem is the same as showing that there exist integers $x$ such that

$$P(x)^2 \not \equiv P(x) \pmod{10}.$$

For the time being, I propose a bad solution. I believe that there exists a much simpler, beautiful solution to this:

Let me consider few cases:

$\textbf{Case 1.}$ Suppose $n\equiv 0, 1,3,5, 6, 8 \pmod{10}$. Then, consider $x\equiv 9 \pmod {10}$. We can easily see that

$$P(x) \equiv 9m+n+9p+1 \equiv 9m+9n+9p+2n+1 \pmod{10}.$$

That is

$$P(x) \equiv 9(m+n+p) + (2n+1) \equiv 1+(2n+1) \pmod{10} \equiv 2n+2 \pmod {10}.$$

Here I have used your observation that $m+n+p\equiv 9\pmod{10}$. Clearly for the given condition on $n$, we can easily infer that

$$P(x) \not\equiv 0, 1, 5, 6 \pmod{10}.$$

Thus $$P^2(x) \not \equiv P(x) \pmod {10},$$ in this case.

$\textbf{Case 2.}$ Suppose $n\equiv 2, 7 \pmod{10}$. We now proceed with two cases based on the nature of $m$. Suppose $m\equiv 1,2,3,6,7,8 \pmod{10}$. Consider $x\equiv 3 \pmod {10}$. We, then would have

$$P(x) \equiv 7m+9n+3p+1 \equiv 3(m+n+p)+4m+6n+1 \pmod{10}.$$

That is

$$P(x) \equiv 4m+6n+8 \pmod{10},$$ where we have used $m+n+p \equiv 9 \pmod {10}$. Now, for the given condition on $n$, we can easily see that

$$P(x) \equiv 4m \pmod{10}.$$

Now, for the values of $m$ chosen, we have

$$P(x) \not \equiv 0, 1, 5, 6 \pmod{10}.$$

Thus, in this case $P(x)^2 \not \equiv P(x) \pmod{10}.$

Now, consider the second case, $m\equiv 0,4,5,9 \pmod{10}$. Choose $x\equiv 7 \pmod{10}$. Then

$$P(x) \equiv 3m+9n+7p+1 \equiv 7(m+n+p)+6m+2n+1 \equiv 6m+8 \pmod{10},$$ for the given values of $n$. Thus for the given values of $m$, it can be easily seen that

$$P(x) \not\equiv 0,1,5,6 \pmod{10}$$

Hence $P(x)^2 \not \equiv P(x) \pmod{10}$ in this case.

$\textbf{Case 3.}$ Suppose $n\equiv 4, 9 \pmod{10}$. In this case, suppose $m\equiv 0, 3, 4, 5, 8, 9 \pmod{10}$. Choose $x\equiv 3 \pmod{10}$. Then, clearly $$P(x) \equiv 7m+9n+3p+1 \pmod{10} \equiv 3(m+n+p)+4m+6n+1 \pmod{10}.$$

That is

$$P(x) \equiv 4m+6n+8 \pmod{10}.$$

For the given values of $n$, we have

$$P(x) \equiv 4m+2 \not \equiv 0,1,5,6 \pmod{10},$$ for the given values of $m$. Thus, in this case $P(x)^2 \not \equiv P(x) \pmod{10}.$ Suppose $m\equiv 1, 2, 6, 7 \pmod{10}$. Consider $x\equiv 7 \pmod{10}$. Then

$$P(x) \equiv 3m+9n+7p+1 \equiv 7(m+n+p)+6m+2n+1\pmod{10}.$$

Then,

$$P(x) \equiv 6m+2n+4 \equiv 6m+2 \pmod{10},$$ for the given values of $n$. Clearly for the values of $m$ chosen, we will have $P(x) \not\equiv 0,1,5,6\pmod{10}$. Hence, $P(x)^2 \not\equiv P(x) \pmod{10}$.

We have covered all the cases.

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Suppose that $(P(a))^2$ and $P(a)$ have the same units digit in decimal for every integer $a$. Then for all $a$ you have $P(a)\equiv0,1\pmod{5}$, and so every element of $\Bbb{F}_5$ is either a root of $P(x)$ or of $P(x)-1$. Both are nonzero cubics over $\Bbb{F}_5$ because $P(0)\equiv1\pmod{5}$ and $P(1)\equiv0\pmod{5}$. It follow that one of them has three roots and the other has two roots in $\Bbb{F}_5$. This is of course impossible because they are cubics.

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