Counting polynomials with same coefficient sum and a given value at a point Call an univariate polynomial $f(x) = \sum_{i=0}^{n}a_{i}x^{i} \in \Bbb{Z}[x]$ symmetric if $a_{i} = a_{n-i}$ and $a_{0} = a_{n} > 0$. 
For a given $\sum_{i=0}^{n}a_{i}$ and $a_{i} \geq 0$, how many symmetric polynomials $f(x) = \sum_{i=0}^{n}a_{i}x^{i} $ over $\Bbb{Z}[x]$ are there such that for two fixed integers $r,s$ $f(r)=s$? (two cases: first case - the degree $n$ is same for both the polynomials and second case - the degree $n$ may be different for both the polynomials)
Are there more than one such polynomials (if one exists)?
How does one construct any such polynomial?
 A: The polynomials $f_1=x^4+9x^3+9x+1$ and $f_2=3x^4+14x^2+3$ both have coefficient sum $f_1(1)=f_2(1)=20$ and the common value $f_1(2)=f_2(2)=107.$ The key is that the difference $f2-f1=g=2x^4-9x^3+14x^2-9x+2$ is symmetric with $g(1)=g(2)=0.$  we could take $f_2=f_1+g$ for any symmetric degree $4$ polynomial with large enough positive coefficients. We can also multiply both $f_1$ and $f_2$ by any symmetric integer polynomial.
In general
If $f_1(x)$ and $f_2(x)$  have 


*

*equal degree

*equal value $f_1(r)=f_2(r)=s$ and 

*equal coefficient sum $f_1(1)=f_2(1)$ 

*positive coefficients


then the difference $g(x)=f_2(x)-f_1(x)$ will have $g(r)=g(1)=0$. Once we have a suitable $g(x)$ we can build many eamples $f_1,f_2$
So 


*

*start with a symmetric polynomial having a root $r \ne 0$ say $rx^2-(r^2+1)x+r$ 

*multiply that by $x^2-2x+1$ to get a symmetric polynomial, in this case  $g(x)=r{x}^{4}- \left( {r}^{2}+2\,r+1 \right) {x}^{3}+ \left( 2\,{r}^{2}+2 \,r+2 \right) {x}^{2}- \left( {r}^{2}+2\,r+1 \right) x+r,$ 
where $g(r)=0$ and also $g(1)=0$ (i.e. the coeffixcients add to $0$). 

*Now pick another symmetric polynomial $f_1(x)$ of the same degree with positive coefficients and set $s=f_1(r)$. 


Then $f_1$ and $f_2=f_1+g$ will have the first three properties above and will have the fourth property provided the coefficients of $f_1$ are large enough. 
If you relax the conditions slightly, then the $g$ above gives $f_1=9x^3+9x$ and $f_2=2x^4+14x^2+2$ with unequal degrees, coefficient sum $18$, and $f_1(2)=f_2(2)=90.$ 
This does not give an example with unequal degrees yet both with non-zero constant term.

This seems like cheating, but would you consider the polynomials $f_1=x^2+2x+1$ and $f_2=4$ an example with $f(-3)=4$ in both cases? In general $r=-b-1$ and  $s=b+2$ for $f_1=x^2+bx+1$ and $f_2=b+2$
It looks less trivial if we, as mentioned above,  multiply both sides by $x+1$ or some other symmetric polynomial to get things like $f_1=x^3+3x^2+3x+1$ and $f_2=4x+4$ both with $f(-3)=-8.$
A: What you're asking for is related to the classic Frobenius Problem.  Let me illustrate with the example you and Gerry Myerson discussed in comments.
In looking for symmetric polynomials of degree $8$ with non-negative coefficients summing to $\sigma$ (which we'll set shortly to $18$, so that Gerry's polynomial is a solution) for which $f(r)=s$ (soon to be set to $r=3$, $s=14842$), we must have $a(r^8+1)+b(r^7+r)+c(r^6+r^2)+d(r^5+r^3)+er^4=s$ with $2(a+b+c+d)+e=\sigma$.  This can be rewritten as
$$a(r^8-2r^4+1)+b(r^7-2r^4+r^3)+c(r^6-2r^4+r^2)+d(r^5-2r^4+r^3)=s-\sigma r^4.$$
Now plugging in $r=3$, $s=14842$ and $\sigma=18$ (and removing what turns out to be a common factor of 4) produces
$$1600a+507b+144c+27d=3346$$
to be solved in non-negative integers $a$, $b$, $c$, and $d$ (after which one must also check that $e=18-2(a+b+c+d)$ is also non-negative).  You can check that Gerry's coefficients, $(1,3,1,3)$, satisfy this equation.  I'll leave it to someone else to find the rest.
Note that changing $s$ simply changes the number on the right hand side whose sum as a combination of "coins" of value $1600$, $507$, $144$, and $27$ is sought.  (To be precise, $s$ can only change by multiples of $4$ in this example.)  The Frobenius problem looks for the largest number that cannot be so represented, which in general seems to be difficult to decide.  It's possible that the special form of the numbers here (all of the form $r^m-2r^{n/2}+r^{n-m}$) allows for a Frobenius miracle, but if they do I don't see how.  Maybe someone with more knowledge of the Frobenius problem can comment or answer with greater authority.
