What is the smallest element of $\{n\in \mathbb N^*\mid \underbrace{1+...+1}_{n\ times}=0\}$

Question

Let $R$ be a field and consider $$A=\{n\in \mathbb N^*\mid \underbrace{1+...+1}_{n\ times}=0\}.$$ Assuming that it's not empty, prove that its smallest element is prime.

I have no idea how to do it. It looks here that we are in $\mathbb Z/n\mathbb Z$ and since $R$ is a field and that $A$ might be a subfield, then $n$ will be prime, but this don't answer the question since they ask for the smallest element.

If my question is unclear, let me know.

$A$ cannot be a subfield because it is a set of integers, not a subset of the field at all. It is almost an ideal of $\mathbb Z$, except that it lacks the non-positive elements. — hmakholm left over Monica, Sep 25 '16 at 12:12
@HenningMakholm: What could it be except $\mathbb N\backslash {0}$ ? — user352653, Sep 25 '16 at 12:42

score 2 · Answer 1 · answered Sep 25 '16 at 12:06

2

Assume its smallest element $n$ is not prime and write $n=ab$ where $a\geq 2$, $b\geq 2$. Then: $$\underbrace{(\underbrace{1+...+1}_{a\ times})+(\underbrace{1+...+1}_{a\ times})+\cdots+(\underbrace{1+...+1}_{a\ times})}_{b\ times}=0$$ Rewrite this as $$b\cdot(\underbrace{1+...+1}_{a\ times})=0.$$ Can you spot a contradiction?

answered Sep 25 '16 at 12:06

Olivier Moschetta

4,247

Either $b=0$ or $a=0$ ? which contradict $a,b\geq 2$ ? But don't we directly have that $ab=n=0$ which is a contradiction ? – user352653 Sep 25 '16 at 12:34
but $b$ is not an element of $R$ – mercio Sep 25 '16 at 13:39
$b$ is shorthand for $b\cdot 1$ which is well defined. – Olivier Moschetta Sep 25 '16 at 13:46
Actually, writing it as $$(\underbrace{1+\cdots+1}{b\text{ times}})\cdot(\underbrace{1+\cdots+1}{a\text{ times}})=0$$ would be clearer. - @user352653: The originally assumed $n=ab$ is an equation in the integers, not in $R$, so $n\ne 0$ by assumption. – hmakholm left over Monica Sep 25 '16 at 14:09

What is the smallest element of $\{n\in \mathbb N^*\mid \underbrace{1+...+1}_{n\ times}=0\}$

1 Answers1