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Two polynomials are equal, if and only if they have the same coefficients and degrees for the corresponding terms (or they can be so arranged.)

A polynomial equation is when both sides of the equation is a polynomial expression.

These are two different mathematical statements, but strangely, I cannot think of a way to express these two statements with different notations so that the difference is clear.

When we write a polynomial equation, in which both sides of the equal sign is a polynomial expression, this represents an equation to be solved for the variable(s) in question, but does not mean the actual polynomial expressions are equal, the conditions for which I have given above.

If so, how do we express mathematically that two polynomials are equal and that we are not talking about an equation?

Prem
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jacob78
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  • Most of the time, the difference is made sufficiently clear from context. For instance, if you are solving for a length of a line segment in a geometric diagram, you might set that length to be x and write down an equation in terms of x using the known geometric properties in the diagram. In this case, it’s clear an equation is meant, not an identity. – Divide1918 Dec 09 '23 at 05:29
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    In the case of the equality you can write an additional statement “for all real x”. This also generalizes to when the domain is something different like complex numbers. – Jack Dec 09 '23 at 05:37
  • I have included only 2 Example Usage Images in my Post , to shorten the longish Answer , Jacob , though there are other text books & online articles using that. – Prem Dec 09 '23 at 06:02
  • I’m voting to close this question because the first comment answers the question. – Kurt G. Dec 09 '23 at 06:41
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    That may justify moving Comments to Answers , @KurtG. , not deleting Questions itself ! – Prem Dec 09 '23 at 07:12
  • @Prem Closing is not deleting. The matter is trivial and should not attract further formal answers. – Kurt G. Dec 09 '23 at 08:33
  • @Jack That works for infinite fields, but it requires care in other cases. For two polynomials $p,q \in F[x]$ where $F$ is finite, “$p = q$” and “$p(x) = q(x)$ for all $x\in F$” are different. – Eike Schulte Dec 09 '23 at 09:30

2 Answers2

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Notation :

I have seen & used this Notation :

Polynomial Equation : $P_1(x)=P_2(x)$
Eg : $x^2+x=x^2+1$
$x^2+1=2x$

Polynomial Identity [ which OP is calling Polynomial "Equality" ] : $P_1(x) \equiv P_2(x)$
Eg : $x^2+2x+1 \equiv 1+2x+x^2$
$(x+y)^2 \equiv x^2+2xy+y^2$

Usage in text books & articles :

IDENTITY 1
"INTRODUCTORY MATHEMATICS FOR ENGINEERS
A. D. MYSKIS"

IDENTITY 2
https://slideplayer.com/slide/3358130/

OBSERVATIONS :

( This Section was added before I added the Images ! )

This is in line with trigonometry :
Equation : $\sin(x)=\cos(x)+1$
Identity : $\sin(2x) \equiv 2\sin(x)\cos(x)$

This Notation is used when we want to Distinguish the 2 Cases & we want to be rigorous.
Otherwise , most of the time "$=$" will do , where "$\equiv$" might be used. Myskis alluded to the Same Point.

A View-Point is that Equation might have no Solution , unique Solution , infinite Solution , Etc while Identity is some Equation where the whole Domain is the Solution Set !

We read "$=$" like "Equal to"
We could read "$\equiv$" like "Identically Equal to" & "Equivalent to"

Prem
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  • I understand. This leads me to believe we could, in order to avoid confusion, say that two polynomials are "identical" instead of "equal" when we mean that they are the same polynomial and we are not solving for the variable. – jacob78 Dec 09 '23 at 06:03
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    That is Correct , @jacob78 , though in the Penultimate Paragraph , I gave the "Alternate View-Point" that when we try solving the "Identity Equation" , we will get "Solution Set == Whole Domain" ! – Prem Dec 09 '23 at 06:11
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Q: "... how do we express mathematically that two polynomials are equal and that we are not talking about an equation?"

A: Normally, with words.

Usually something like "We know that for all x ..." for the first case, and "Solve ..." or "Find all $x$ such that ..." for the second case. Often there is enough context that it is clear which is intended without being explicit, but to your generic question that's the answer.

If you like being pedantic, you can write

$$ \forall x \,\,\,\,\,p_1(x) = p_2(x) $$ and $$ \{x \mid p_1(x) = p_2(x)\} $$ respectively, but you don't gain much clarity with these. Maybe a computer program might prefer them, but most humans like words.

Edited to add, replying to comment:

Adding the quantifier might resolve everything for a polynomial in one variable, but you might want to think about what's going on if you have functions of two variables, and you are asked for:

  • Find all $y$ such that $p_1(x,y) = p_2(x.y)$ for all values of $x$.

Again, you can encode this with just notation by considering $$\{y \mid \forall x \,\,\,\, p_1(x,y) = p_2(x.y) \}$$

Maybe we can broaden your comment to say that we only have clarity when there are no free variables left. But note that the traditional definition of a free variable is one without a $\forall$ or $\exists$ modifier, and we broaden that to include "set of all $x$ such that ...".

JonathanZ
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  • I think the core point is to realize that polynomials include variables which make an equation of polynomials an "open statement" unlike equations where both sides are constant. Using the universal quantifier to make the distinction resolves the issue. – jacob78 Dec 09 '23 at 05:56
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    @jacob78 I disagree to the extent that simplistic use of a universal quantifier specifies a domain for $x$. And this need not be sufficient for the purposes of identifying a polynomial. It is the old polynomial vs. polynomial function confusion. A univariate polynomial of $x$ is, essentially, the sequence of coefficients of various powers of $x$ as opposed to a function that specifies a range of values for $x$. The standard example is that $f(x)=x^2+x$ is a quadratic polynomial in the ring $\Bbb{Z}_2[x]$, hence distinct from $g(x)=0$. – Jyrki Lahtonen Dec 09 '23 at 06:33
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    (cont'd) Even though we have $f(a)=g(a)$ for all $a\in\Bbb{Z}_2$. True, with real (or complex, or rational) numbers this distinction is moot. But only because we have a theorem saying so! – Jyrki Lahtonen Dec 09 '23 at 06:35