Can we tile $5\times 5$ table with tiles $2\times 3$ so that each $1\times 1$ cell of the table is covered with the same number of tiles? 
Can we tile $5\times 5$ table with tiles $2\times 3$ so that each $1\times 1$ cell of the table is covered with the same number of tiles?

We assign to cell $(i,j)$ value $z^{i+j}$ where $i,j\in \{0,1,2,3,4\}$. So if it is possible then exists natural number $n$ and a polynomial $q$ of degree $5$ such that:
$$(1+z+z^2+z^3+z^4)^2 \cdot n = (1+z)(1+z+z^2)q(z)$$
Since this must be true for all $z$ it must true also for $z=-1$, so we get $n=0$, so it is impossible.
Have I missed something here?
 A: Your answer is fine.
A more elementary answer is that $i+j$ is even for $13$ squares and odd for $12$ square, but the tile covers three even and three odd squares.

Your argument is nice because it generalizes to more cases. My argument shows that a rectangular tile with an even number of elements cannot cover evenly a rectangle with an odd number of squares.
Yours is stronger. If $m\times n$ is your board, and $a\times b$ is the size of the tiles, you’d need an integer polynomial $q(x)$ and non-zero integer $N$ so that:
$$N(x^m-1)(x^n-1)=(x^a-1)(x^b-1)q(x)\tag{1}$$
But by unique factorization in $\mathbb Z[x],$ if there is a solution, there is a solution with $N=1.$
Of course, we need a $q$ with positive coefficients, but if $N>1,$ then $N$ is a factor of $q$ and $q/N$ still has positive coefficients. This means if there is a even tiling, there is a tiling in the usual sense.
Now, all the roots on the right side of (1) must be roots on the left side, so either$a\mid m$ or $a\mid n,$ and either $b\mid m$ or $b\mid n.$
And any repeated root on the right side, a root of $x^{\gcd(a,b)}-1,$ must be a root of $x^{\gcd(m,n)}-1,$ so $\gcd(a,b)\mid \gcd(m,n).$
If $m=n,$ this means $a,b\mid m,$ and clearly this is sufficient.
If $a\mid m$ and $b\mid n$ or vice versa, then we can also clearly tile.

We don’t always get that $a\mid m,b\mid n$ or vice vesa. For example, we can tile a $6\times 5$ rectangle with $2\times 3$ tiles. We can obviously tile an $6\times 2$ rectangle this way, and a $6\times 3$ rectangle this way. The glue them together. This corresponds to:
$$\begin{align}(x^6-1)(x^5-1)&=(x^6-1)\left[(x^3-1)+x^3(x^2-1)\right]\\
&=(x^2-1)(x^3-1)\left[(1+x^2+x^4)+x^3(1+x^3)\right]
\end{align} $$
If $a,b\mid m$ and neither $a,b\mid n,$ we can still tile if $$n=ax+by$$ for positive integers $x,y.$
I suspect these are all the cases, but I haven’t proved it.

Of course, this polynomial greatly eliminates some of the tiling information.
We have an even tiling if and only if there are integer polynomials $q_1(x,y),q_2(x,y)$ with no negative coefficients and an integer $N>0$ so that:
$$
\begin{align} 
f(x,y)&=N(x^m-1)(y^n-1)\\&=(x^a-1)(y^b-1)q_1(x,y)+(x^b-1)(y^a-1)q_2(x,y).\tag{1}
\end{align} 
$$
Not sure if this formulation helps much.
We can certainly use it to solve the case where $a=b,$ because this means that $(x^a-1)(y^a-1)$ is a divisor of $(x^m-1)(y^n-1),$ so $a\mid m,n.$
Also if $a\mid b.$ Then the right side of (1) is divisible by $(x^a-1)(y^a-1),$ so we again get $a\mid m$ and $a\mid n.$ We already know $b$ divides one of $m,n.$

My original argument can be altered to show these results.
Color each square in the rectangle with $i+j\pmod a.$ Then an $a\times b$ rectangle will, for each $i=0,\dots,a-1,$ will cover exactly $b$ squares with cover $i.$ So, to have a tiling, we need the same number of squares for each color $i.$
If neither $m$ nor $n$ is divisible by $a,$ I think we can show that the coloring isn’t even. If $m=aq_1+r_1,$ and $n=aq_2+r_2,$ with $0<r_i<a,$ we know that the rectangle $aq_2\times n$ and rectangle $r_1\times aq_2$ also. This is left with a rectangle of size $r_1\times r_2.$
It might take polynomial approach to show this is uneven. Basically, if $\zeta$ is a primitive $a$th root of $1,$ you’d need $$(1+\zeta+\cdots+\zeta^{r_1-1})(1+\zeta+\cdots+\zeta^{r_2-1})=0$$ which is impossible with $0<r_i<a.$
A: What you are arguing, somewhat indirectly, is that if you colour the table in chequerboard fashion, then each $2\times3$ tile covers as many dark as light squares. Since the entire board has a colour imbalance, the only way to cover it entirely precisely $n$ times (even allowing for negative multiplicities of certain tile positions) is to have $n=0$. Since it is definitely possible to have $n=0$ (using no tiles at all), I would say that answer to the question is strictly speaking "yes", unless one construes the used verb "tile" to somehow implicitly exclude this trivial solution.
