${{\left( {{a}_{n}} \right)}_{n\in {{\mathbb{N}}^{*}}}}\subset {{\mathbb{N}}^{*}}$ is called complete,if $\forall n\in {{\mathbb{N}}^{*}},\exists p,q\in {{\mathbb{N}}^{*}}$ such that $np={{a}_{q}}$. Let $a,r\in {{\mathbb{N}}^{*}}$.Prove that ${{\left( a+\left( n-1 \right)r \right)}_{n\in {{\mathbb{N}}^{*}}}}$ is complete$\Leftrightarrow r\left| a \right.$

First of all,thanks ion for giving this beautiful problem.And secondly,please point out any flaw in my proof. First of all I prove that if {a+(n-1)r} is a complete sequence,then a divide r. It is easy to prove.Take n=r and define An=a+(n-1)r.Then the condition is satisfied iff r divides Aq for some q. Now for any q r divides Aq iff r divdes a.So the first one is proved.Now my task is to prove the converse. Let r divides a and a=kr for some integer k.Thus Aq=a+(q-1)r=r(k+q-1) for any q. .Let for some integer n there exist p,q such np=Aq(due to my inability to write in LaTeX please denote the An's and Aq's as suffixes)Now I am fixing p=r as I can take the integers p and q as my wish.So the thing stands like nr=r(k+q-1) or, n=k+q-1 or, q=n+1-k.Thus for any n I can find q according to this relation such that np=Aq.Thus {Aq=a+(q-1)r} is a complete sequance.Hence,my proof.I again say please report any flaw in the proof.Good Bye and keep posting problems.

Hey someone please reply about the genuinity of the proof.It is urgent

Don't bump for 3 hours. mathdebam wrote: First of all I prove that if {a+(n-1)r} is a complete sequence,then a divide r. It is easy to prove.Take n=r and define An=a+(n-1)r.Then the condition is satisfied iff r divides Aq for some q. Now for any q r divides Aq iff r divdes a.So the first one is proved. I'd say you have the right idea, but you write it incorrectly. I'd write something like this: Suppose that for the sake of contradiction $\{a + (n-1)r \}$ is complete but $r$ doesn't divide $a$. Note that $a + (n-1)r \equiv a \bmod r$ for all $n$; since $r$ doesn't divide $a$, we have $a \not\equiv 0 \bmod r$. But this means no element in the sequence is divisible by $r$, contradicting the fact that the sequence is complete. mathdebam wrote: Let r divides a and a=kr for some integer k.Thus Aq=a+(q-1)r=r(k+q-1) for any q. .Let for some integer n there exist p,q such np=Aq(due to my inability to write in LaTeX please denote the An's and Aq's as suffixes)Now I am fixing p=r as I can take the integers p and q as my wish.So the thing stands like nr=r(k+q-1) or, n=k+q-1 or, q=n+1-k.Thus for any n I can find q according to this relation such that np=Aq. You also have the right idea, but you fail to handle small cases. If you take $p = r$, it's possible that $q \le 0$ which is invalid (because the sequence is only defined for positive indices). You should divide to two cases where $n < k$ and $n \le k$; only for the latter that you can use $p = r$. For the former case, I'd take $q = k! - k + 1$, so $np = A_q = rk!$. Since $n < k$, there always exists $p$. In addition, since $k! \ge k$ for all $k$, $q \ge 1$, so $A_q$ is always defined. Well, your proof has the right idea, but it's a bit awkwardly phrased (not to mention readability issues due to lack of LaTeX), and it misses some cases.

I think that it is not undefined for negative natural numbers because in the question it is mentioned that the sequence will contain non-zero terms .No harm in having negatives,isn't it?And if the sequence can contain negative integers aand if you define Aq=a+(q-1)r,then what is the harm in having non-negative indices.

ionbursuc wrote: ${{\left( {{a}_{n}} \right)}_{n\in {{\mathbb{N}}^{*}}}}\subset {{\mathbb{N}}^{*}}$ ionbursuc wrote: ${{\left( a+\left( n-1 \right)r \right)}_{n\in {{\mathbb{N}}^{*}}}}$ Everywhere ionbursuc says $\mathbb{N}^*$, which means either $\{0,1,2,\ldots\}$ or $\{1,2,3,\ldots\}$; either way, negatives don't count. Also, having negative indices don't do anything bad, but you can't then say "the sequence is complete because $4$ divides $a_{-1}$", because the definition of a complete sequence says that a number divides a term in the sequence ${\{a_n}_{n \in \mathbb{N}^*}$; aka, divides a term with nonnegative index. It doesn't matter whether a number divides a negative-index term; the thing is, to be complete, the number must divide a nonnegative-index term. (Unless you can prove that if a number divides a negative-index term then the number also divides some nonnegative-index term, so you can chain this with the condition of completeness, but you haven't done it.)

No.Actually I think that the rewording of ion is wrong.He should not have used N* as in the original problem there is no mention of the word positive or non-negative.It is always non-zero.Actualy it is the problem of the Romanian National Olymp[iad 2013 .You can actually check it from yhe contest site.

$\mathbb{N}=\left\{ 0,1,2,... \right\}$ and ${{\mathbb{N}}^{*}}=\left\{ 1,2,... \right\}$

This problem is very strange but easy. Suppose that $ {{( a+( n-1)r)}_{n\in{{\mathbb{N}}^{*}}}} $ is complete. Then $\exists n \in \mathbb{N}$ such that $r|a+(n-1)r \implies r|a$. Take $a=rm$. Then $ {{( a+( n-1)r)}_{n\in{{\mathbb{N}}^{*}}}} = {{( r(m+n-1) }}$. Obviously, $\forall p \in \mathbb{P}, x \in \mathbb{N}, \exists n \in \mathbb{N}$ such that $p^x|m+n-1$. Now, $\alpha|m+n_1 -1$ and $\beta|m+ n_2 -1 \implies \alpha \beta|m+[m+(m+n_1+n_2-3)+2]-1$ and by induction, we are done.

Let $(a_n)_{n\geq 1}$ an arithmetic progression we denote as $r$ the common difference. Obviously, $r\neq 0.$ If $r|a_1$ then there exists some $d\in\mathbb{N}$ such that $a_1=rd,$ then we can easily see that $a_n=r(d+n-1)$ and now we just have to choose $(d+n-1)$ such that $(d+n-1)$ is a multiple of some $k\in\mathbb{N}.$ Now, we will prove $r$ must divide $a_1$ if $(a_n)_{n\geq 1}$ is an complete arithmetic progression. Because any nonzero integer has at least one multiple in an complete series, $r$ will also have at least one multiple in $(a_n)_{n\geq 1},$ lets call it $a_p,$ where $p\in\mathbb{N},$ then $a_p=a_1+(p-1)r$ but $r|a_p=a_1+(p-1)r$ which implies $r|a_1$ and the conclusion follows.