(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

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?

Answer 1

The second solution is not so wrong

$a^5-1=0$ has five solutions, the five complex fifth roots of unity. Namely

$$\left(a_0=1,\;a_1 = -\frac{1}{4}-\frac{\sqrt{5}}{4}-i \sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}},\;a_2 = -\frac{1}{4}+\frac{\sqrt{5}}{4}+i \sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}},\\a_3 = -\frac{1}{4}+\frac{\sqrt{5}}{4}-i \sqrt{\frac{5}{8}+\frac{\sqrt{5}}{8}},\;a_4 = -\frac{1}{4}-\frac{\sqrt{5}}{4}+i \sqrt{\frac{5}{8}-\frac{\sqrt{5}}{8}}\right)$$

One of them, $a=1$, is not a root of the fourth degree equation $$a^4+a^3+a^2+a+1=0$$ but the other four actually are because $$a^5-1=(a-1)(a^4+a^3+a^2+a+1)$$

So plugging $a^5=1$ in the $a^{2000}+a^{2010}+1$ is perfectly legal and the actual result is $3$

Hope this is useful

Answer 2

The first solution is wrong, because $3\neq z!$ for any $z$.

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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 $a=1$.