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I wonder if the following is possible: Let $k$ be a field of characteristic different from $2$.

Can the polynomial $x(x-1)(x-2)$ become the square of another polynomial by evaluating it in another polynomial?

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  • $\begingroup$ Â remark (I don't know if it is useful) : writing your polynomial under the form $(x-1)((x-1)^2-1)$ your issue is the same as writing $X(X^2-1)$ under the form $P(X)^2$ $\endgroup$
    – Jean Marie
    Dec 4, 2022 at 21:50
  • $\begingroup$ What do you mean by "evaluating it in another polynomial"? If you mean finding a polynomial $P$ such that $P((x(x-1)(x-2))$ is the square of another polynomial, then all you have to do is take $P$ to be the square of a polynomial, e.g., $P(x)=x^4$. $\endgroup$ Dec 4, 2022 at 22:45
  • $\begingroup$ @GerryMyerson I mean finding a polynomial $P$ such that $P(P-1)(P-2)$ is the square of another polynomial $\endgroup$ Dec 4, 2022 at 23:10

2 Answers 2

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Suppose $f,g\in k[x]$ satisfy $f(f-1)(f-2)=g^2$. Then by unique factorization we have $f$ divides $g$, so $f^2$ divides $f(f-1)(f-2)$, so $f$ divides $(f-1)(f-2)=f^2-3f+2$, so $f$ divides 2. Thus $f$ is a unit (nonzero scalar).


Edit. The above argument is not correct. Here is a new argument.

We know that $f$, $f-1$ and $f-2$ are pairwise coprime. Let $p$ be an irreducible factor of $f$. Then $p$ divides $g$, so $p^2$ divides $g^2$, and so $p^2$ divides $f$. Cancelling $p^2$ and repeating shows that $f=a^2$ is itself a square. Similarly $f-1=b^2$ is a square. Now $$ 1 = a^2-b^2 = (a+b)(a-b) $$ has degree zero, so $a\pm b$ are both scalars, and hence $2a$ is a scalar. Since $2\neq0$ in our field $k$, we deduce that $a$, and hence $f$, is a scalar.

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  • $\begingroup$ How do you use unique factorization to ensure that $f$ divides $g$.? Can it not happen that $f$ has a root of $g$ with multiplicity 1 in $g$ and multiplicity $2$ in $f$? $\endgroup$ Dec 6, 2022 at 21:45
  • $\begingroup$ You’re right, I was too hasty. I have now added a new argument. $\endgroup$ Dec 6, 2022 at 22:58
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It is not possible for the polynomial $x(x-1)(x-2)$ to become a perfect square by evaluating it in another polynomial. To see why, note that the polynomial $x(x-1)(x-2)$ has degree 3, which is odd. This means that it cannot be expressed as the square of another polynomial, which would have degree 2 or even.

For example, if we try to write $x(x-1)(x-2)$ as the square of a polynomial $f(x)$, we would have

$$f(x)^2 = x(x-1)(x-2)$$

Expanding the right-hand side, we get

$$f(x)^2 = x^3 - 3x^2 + 2x$$

Comparing the degrees of the two sides, we see that the left-hand side has degree 2, while the right-hand side has degree 3, which is a contradiction. This means that the polynomial $x(x-1)(x-2)$ cannot be written as the square of another polynomial.

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  • $\begingroup$ But if I evaluate it in a polynomial of even degree (e.g. $x^2$), the polynomial become of even degree: $x^6 -3x^4+2x^2$ $\endgroup$ Dec 4, 2022 at 21:20
  • $\begingroup$ What is true with $k=\mathbb{R}$ may not be true in a general field. $\endgroup$
    – Jean Marie
    Dec 4, 2022 at 21:44
  • $\begingroup$ While it is true that evaluating the polynomial $x(x-1)(x-2)$ at $x^2$ would result in a polynomial of even degree, this does not mean that the original polynomial $x(x-1)(x-2)$ can be the square of another polynomial. $\endgroup$
    – Nuro007
    Dec 4, 2022 at 21:55
  • $\begingroup$ When you evaluate a polynomial $p(x)$ at a value $a$, you are essentially replacing all occurrences of the variable $x$ in $p(x)$ with the value $a$. So, evaluating the polynomial $x(x-1)(x-2)$ at $x^2$ would result in the polynomial $(x^2)(x^2-1)(x^2-2)$, which does have an even degree. However, this does not change the fact that the original polynomial $x(x-1)(x-2)$ has an odd degree and cannot be the square of another polynomial. $\endgroup$
    – Nuro007
    Dec 4, 2022 at 21:55

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