<s>[INST] (1) Sum of two factorials in two ways; (2) Value of $a^{2010}+a^{2010}+1$ given $a^4+a^3+a^2+a+1=0$. Question $1$:
Does there exist an integer $z$ that can be written in two different ways as $z=x!+y!$,where $x,y\in \mathbb N$ and $x\leq y$?

Answer: $0!=1!$ so $0!+2!=3=1!+2!$

Question $2$:
If $a^4+a^3+a^2+a+1=0$, find the  value of $a^{2010}+a^{2010}+1$.

Answer: If $a^4+a^3+a^2+a+1=0$ then $$(a-1)(a^4+a^3+a^2+a+1)=0\implies a^5-1=0\implies a=1$$So the value of the required expression is $3$

I think that it is wrong as  at $a=1$,$\frac{a^5-1}{a-1}$ is not defined.
Please give me some hints  to solve thes problems?
 [/INST] The first solution is wrong, because $3\neq z!$ for any $z$.

The second solution is wrong.
If $a=1$, then $a^4+a^3+a^2+a+1=1+1+1+1+1\neq 0$, so it is obvious you made a serious mistake in your second solution.
In particular, you multiplied the equation by $(a-1)$, which of course produced a solution of $a=1$.
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