# On the existence of certain type of arithmetic progressions

Suppose that we talk here only about arithmetic progressions whose terms are natural numbers and whose first term is $a_1=1$ and which have at least three terms and whose common difference $d$ is a natural number.

I observed this for some small values of $S$.

If we choose $S=9$ then there exists arithmetic progression $a_n$ such that the sum of at least three terms (starting from the first term) is $9$. We have $a_1=1$ and we can take $d=2$ and $n=3$ to obtain $a_1+a_2+a_3=1+3+5=9$.

If we choose $S=15$ then there is more than one solution, we have $a_1=1$ and we can take $d=4$ and $n=3$ to obtain $a_1+a_2+a_3=1+5+9=15$ and we have $a_1=1$ and we can take $d=1$ and $n=5$ to obtain $a_1+a_2+a_3+a_4+a_5=1+2+3+4+5=15$.

If we choose $S=8$ then there is no solution because $1+2+3$ is not a solution and $1+2+3+4$ is not a solution and $1+3+5$ is not a solution and there are no arithmetic progressions of described type "between" those three.

Be aware that if we did not restricts ourselves to arithmetic progressions that have at least three terms then we would always have solution for any natural $S$ because we could choose arithmetic progression with two terms that are $a_1=1$ and $a_2=S-1$ if $S \neq 1$ and one term sequence $a_1=1$ if $S=1$:

So it is quite natural to ask:

Is there a way to determine for which natural numbers $S$ there exists at least one arithmetic progression whose terms are natural numbers and whose first term is $a_1=1$ and which have at least three terms and whose common difference $d$ is a natural number and whose sum of at least three terms (starting from the first) is equal to $S$?

• The term "natural number" is a bit confusing, as some people consider $0\in\mathbb N$. I'd consider using "positive integer", as otherwise, $a_i=1$ for all $1\le i\le S$ gives a trivial solution. – Emre Jun 18 '16 at 22:41

## This is not a full solution, but rather a demonstration that an answer to this question would be long.

Let $$S=\sum_{i=1}^n a_i=\sum_{i=1}^n 1+d(i-1)=n+d\left(\frac {n(n-1)}{2}\right)=\frac {n(2+d(n-1))}2$$

Clearly, $S$ cannot be a prime number, as $d(n-1)+2>n>2$.

For $d=2k$:

• Put $n-1=m$. Then, $S=(m+1)(km+1)$. For example, by putting $k=1$, $S$ can be any perfect-square.

For fixed $d$:

• $S$ can be written as required iff $$8dS+(d-2)^2$$ is a perfect square.

For $n=3$:

• $S=3(d+1)$. So, anything that is divisible by $3$ is okay, except $3$ obviously.

For $n=4$:

• $S=2(3d+2)$. So, anything that is $\equiv 4\pmod6$ is okay, except $4$ obviously.

For $n=5$:

• $S=5(2d+1)$. So, any odd integer that is divisible by $5$ is okay, except $5$ obviously.

For fixed $n$:

• $S\equiv n\pmod {{n(n-1)\over2}}$ gives the solution.

A global analysis seems hard, as the answer will get pretty messy.

For example, If $S=2p$ for some odd prime numbers $p$, then, $n|4p$ and $n<\frac{4p}n$ gives $$p\equiv 2\pmod3$$ or $$p=3$$

If $S=pq$ for some odd prime numbers $p<q$, then, $n|2pq$ and $n<\frac{2pq}n$ gives $${p-1\over 2}\lvert q-1$$ or $$2p-1|q-2$$

If $S$ is a power of $2$, then we need $n=2^k$, $k>1$. In order the second term to be even, $d$ needs to be even. Thus, $$S=2^k(1+\frac d2(2^k-1))$$ , suppose the second term is $2^m$ for some $m$. Then, $2^k-1\lvert 2^m-1$. Thus, $m=kl$, $l>1$. So, $S=2^{k(l+1)}$. Thus, the exponent of $S$ cannot be prime, but every other exponent is fine.

My suggestion would be to restrict the question in some sense, with an extra condition on $n,d,S$.