Equal polynomials of equal degree at equal x must have equal coefficients? does anyone know if the following is true?
I'm looking at something like
$$2^{s_1+s_2+\Delta_2}+2^{s_2+\Delta_2}3-2^{s_1}3^2=2^{s_1+s_2+\Delta_1}+2^{s_2}3-2^{s_1+\Delta_1}3^2$$
And what I notice is that these look like two polynomials (quadratic) at the same $x$ value (if $x=3$) are equal, have the same degree and so I would like to say therefore their coefficients must be equal.
In general maybe
$$2^{s_1+s_2+\Delta_2}+2^{s_2+\Delta_2}x-2^{s_1}x^2=2^{s_1+s_2+\Delta_1}+2^{s_2}x-2^{s_1+\Delta_1}x^2$$
I don't think it works for any $x$, because any two graphs could be equal at an intersection, but maybe the argument works because $3$ is coprime to $2$? 
Is it safe to say that
$$\therefore s_1+s_2+\Delta_2= s_1+s_2+\Delta_1$$
$$\therefore s_2+\Delta_2= s_2$$
$$\therefore s_1= s_1+\Delta_1$$
and so at the end of the day
$$\Delta_1=\Delta_2=0$$
because that result would really make me scream with joy.
Also, could this same argument be applied for similar polynomials of higher degree, maybe something like
$$2^{s_2+s_3+\Delta_3}+2^{s_1+s_3+\Delta_3}3+2^{s_3+\Delta_3}3^2-2^{s_2}3^3-2^{s_1}3^4$$ $$=2^{s_2+s_3+\Delta_2}+2^{s_1+s_3+\Delta_1}3+2^{s_3}3^2-2^{s_2+\Delta_2}3^3-2^{s_1+\Delta_1}3^4$$
allow me to conclude the similar result that
$$\therefore \Delta_1=\Delta_2=\Delta_3=0\mbox{ ?}$$
Any help will be greatly appreciated.
         Edit:

Does this work for $s_1,s_2,\Delta_1,\Delta_2\in\mathbb{R}$? if not,
Does this work if either $s_n\in\mathbb{N}$ or $(s_n+\Delta_n)\in\mathbb{N}$?
 A: Suppose $s_1 = s_2 = 0$, then the equation can be simplified to:
$$
8*2^{\Delta_1} + 4 * 2^{\Delta_2} = 12
$$
Clearly we have the solution of $\Delta_1 = \Delta_2 = 0$.
But, if we substitute $x=2^{\Delta_1},y=2^{\Delta_2}$, then we have a linear equation in two unknowns:
$$
8x + 4y = 12
$$
This equation gives us a line of solutions in the $x,y$-plane. In particular it has infinite solutions where $x,y$ are positive so they can be written as powers of two. 
For example, we can see that $x=\frac{1}{2}, y=2$ is a solution corresponding to $\Delta_1 = -1, \Delta_2 = 1$.
So, even if we assume that $s_1=s_2=0$, there will still be infinite solutions for $\Delta_1, \Delta_2$. 
The situation will be similar for other choices of $s_1,s_2$.
A: If they can be any real numbers it doesn't work, consider the following:
     Set 

$s_2=2\log_23\mbox{ ; }s_1=\log_23\mbox{ ; }\Delta_2=0$.
Then:
$$2^{3\log_23}+2^{2\log_23}3-2^{log_23}3^2=2^{3\log_23+\Delta_1}+2^{2\log_23}3-2^{log_23+\Delta_1}3^2$$
$$3^{3}+3^{3}-3^{3}=2^{\Delta_1}3^3+3^3-2^{\Delta_1}3^3$$
$$0=0$$
Therefore to any $\Delta_1$ the equation is satisfied.
