# Substitution to obtain the square of a polynomial

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?

• 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 $P(X)^2$ Dec 4, 2022 at 21:50
• 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., $P(x)=x^4$. Dec 4, 2022 at 22:45
• @GerryMyerson I mean finding a polynomial $P$ such that $P(P-1)(P-2)$ is the square of another polynomial Dec 4, 2022 at 23:10

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.

• 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 $f$? Dec 6, 2022 at 21:45
• You’re right, I was too hasty. I have now added a new argument. Dec 6, 2022 at 22:58

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.

• But if I evaluate it in a polynomial of even degree (e.g. $x^2$), the polynomial become of even degree: $x^6 -3x^4+2x^2$ Dec 4, 2022 at 21:20
• What is true with $k=\mathbb{R}$ may not be true in a general field. Dec 4, 2022 at 21:44
• 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. Dec 4, 2022 at 21:55
• 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. Dec 4, 2022 at 21:55