Show that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$ cannot be a square when $x\neq3$ and $x\neq2$ respectively. As the title says, I need help showing that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$  cannot be a square when $x\neq3$  and $x\neq2$  respectively, where $x$ is a natural number.
In order to do this, I have to use the fact that if $x$ is any natural number >1, any natural number $n\geq x$ can be represented uniquely in the form
$n=c_0+c_1x+c_2x^2+...+c_mx^m$,
where $0\leq c_i \leq x-1$ for $i=0,1,2,...,m-1$  and
$0< c_m \leq x-1$.

If $x > 3$ and $n = x^4+x^3+x^2+x+1$ is a square then $4n$ is also a square.  However
$$ \begin{eqnarray}4(x^4+x^3+x^2+x+1) &=& (2x^2 + x)^2 + 3x^2 + 4x+4 \\ &=& (2x^2+x+1)^2 - (x+1)(x-3) \end{eqnarray}$$
which shows that
$$(2x^2 + x)^2 < 4n < (2x^2+x+1)^2.$$
Since $4n$ is between two consecutive squares it cannot be a square itself.  In the same way
$$ \begin{eqnarray}x^4+2x^3+2x^2+2x+5 &=& (x^2 + x)^2 + x^2 + 2x+5 \\ &=& (x^2+x+1)^2 - (x+2)(x-2) \end{eqnarray}$$
which shows that $x^4+2x^3+2x^2+2x+5$ is not a square when $x>2$.
