>It’s asking, what is the set of values that make the statement true?
For this to be the case, there would need to be a statement in the first place. And that statement would involve the =, so you necessarily still have the = symbol representing something other than questioness. I would further say that the equation itself is still just a statement, and any "question" interpretation is based entirely on the context where the equation is presented.
Further, lets take seriously the notion that "x+1" is a polynomial in the formal sense. What does it mean to find the set of values for which x+1=2 is true? Normally, I would say that we are looking for the set of x values which makes that equation true. However, we are insisting that the LHS is a polynomial (again, in the formal sense). This means that there is no variable. The "x" in the LHS is literally a value. It makes no sense to ask what values of (0,1,0,0,...) make that equation true. The fact that we give (0,1,0,...) a standard name of x does not suddenly make the question sensical. Nor does the fact x is often used to represent variables.
>I know of no mathematician who thinks x^2-x+1=0 is anything other than a polynomial equation.
To be clear, outside of very particular contexts I would still call x^2-x+1=0 a polynomial equation, because it is extremly useful to talk about polynomials without invoking all of the machinery of formal polynomials.
>And viewing x+3 as an element of C(R) the only reasonable interpretation of x+3=1 is that we are finding the pre-image of 1.
I disagree. Viewing x+3 as an element of C(R), the only reasonable interpretation of x+3=1 is the statement (x↦x+3)=(x↦1).
Viewing x+3 as an element of R would allow us to treat the equation x+3=1 in the "obvious" way. We can prove that the statement x+3=1 implies that x=-2.
Further, I would agree with you that, absent other context, when given an equation which contains an "x" in it, there is some implication that we are supposed to solve for x.
>Your view of how to interpret x^3+4x is too simplistic because the only to way to algebraically manipulate that object is by considering it as an R[x] or R(x).
Why? Suppose I don't know what R[x] or R(x) is. We certainly don't teach highschoolers what either of those are, and they seem to be able to do "algebra" just fine.
Here is a simple approach to dealing with x^3+4x without considering it a member of R[x]. For concreteness, I want to solve x^3+4x=0.
Suppose x \in R such that x^3 +4x = 0.
By the distributive property, this equation is true iff x(x^2 +4x)=0.
By direct calculation, we can verify that x=0 is consistent with this equation, and therefore consistent with the original equation.
Consider the case where x != 0.
Note that the function f(n) = n/x is a bijection. Therefore, we have x(x^2 + 4)=0 iff f(x(x^2+4)) = f(0).
By direct computation, we get that this is true iff x^2+4 = 0.
We know that x^2 >=0, and 4>0.
Therefore, x^2+4 > 0.
This is a contradiction, which means that the case where x!=0 is impossible.
This means that we have proven that x=0.
Now, suppose we were working over C.
Continuing from x^2 + 4 = 0, we can show that:
x^2+4 = 0 iff
(x+2i)(x-2i) = 0 iff
x+2i = 0 OR x-2i = 0 iff
x=2i OR x=-2i
Since the cases x=0 and x!=0 are exhaustive, we have proven the statement x \in {0, 2i, -2i}.
I solved this using the method we teach school children and without invoking any notion of polynomials.