The answer is all perfect squares. Firstly, observe that $d$ works iff $dx^2$ works. Therefore, it suffices to find all squarefree positive integers $d$ which work. If $p \ge 2$ divides $d$, then we'll claim that there doesn't exist such a positive integer $k$. If $p = 2$, then simply note that if $a_1, a_2, \cdots, a_k$ were a permutation of $1, 2, \cdots, k$ which worked, then $a_j+a_{j+1}$ must be even for all $j$ and so the $a_i$'s are of the same parity, which is clearly impossible when $k \ge 3.$ For $p \ge 3$, a similar statement holds that $p|a_j + a_{j+1}$ for all $j$, and so either $p$ divides all of the $a_i$'s or none of them, implying that $k < p$. Hence, $p = 3$ fails as $k \ge 3$. When $p \ge 5$, if we let $a_t = 1$, then we can easily see that all $a_i$'s are $\equiv \pm 1$ (mod $p$), also impossible. Now, it suffices to construct an example when $d = 1$. The following suffices: $$17, 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16,$$ and so the answer is all perfect squares. $\square$

Pathological wrote: then $a_j+a_{j+1}$ must be even for all $j$ and so the $a_i$'s are of the same parity I don't think this is correct... this simply means that the parity is the same for every even and every odd term. I think the argument can be continued, but I'm not sure...

Well we’re assuming that $d$ is squarefree, so in your case that would make $(a_i+a_{i+1})d$ congruent to $2$ mod $4$, clearly not a square.

I don't understand how you are defining the $a_i$...

benstein wrote: I don't understand how you are defining the $a_i$... Pathological wrote: If $p = 2$, then simply note that if $a_1, a_2, \cdots, a_k$ were a permutation of $1, 2, \cdots, k$ which worked What I mean by "worked" here is that $a_1d, a_2d, a_3d, \cdots, a_kd$ is a permutation of $d, 2d, \cdots, kd$ for which the sum of adjacent terms is always a perfect square.

Oh, I understand now, thanks for clarifying. That was not immediately clear...

First observation is that, if $d$ works then $dx^2$ also works.Hence WLOG $d$ is square-free. Now,assume we have $ld$ to the right (or left) of $d$.Then,$l \equiv -1 \pmod{d}$. Again assume $md$ is to the right $(k \ge 3)$ .Then, $m \equiv +1 \pmod{d}$. Hence its easy to see that only $-1,1 \pmod{d}$ appears in coefficients of $d$ in this sequence.But unless $d=1$ thats impossible i.e. that doesn't satisfy our problem condition. Hence,the only possible option left is when $d$ is a perfect square i.e. its square free part is $1$. The construction is given by the user pathological in above post.

The answer is all square numbers. Notice the degree of freedom in the sense of scaling. This means that we can reduce our argument onto all integers $d$ such that $v_p(d) \leq 1$, for any prime $p$. Let $a_1,a_2,\dots ,a_k$ be a permutation of the numbers $1,2,\dots, k$, such that $d\left( a_i + a_{i+1} \right) = t_i^2$, where $t_i$ is an integer. Let $p$ be a prime number such that $p \mid d$, this implies that for every $i$ we must have that $a_i+a_{i+1} \equiv 0 \pmod{p}$. But this implies the following: $$a_1 \equiv a_3 \equiv \dots \equiv a_{1+2t} \pmod{p}$$$$a_2 \equiv a_4 \equiv \dots \equiv a_{2t} \pmod{p}$$For any $t$ as long as $2t+1 \leq k$. But this practically impossible since we must have the numbers $1,2,3$ (since $k \geq 3$). Meaning that $d=1$. Now we just find the construction when $d=1$. Notice that the sequence: $$17,8,1,15,10,6,3,13,12,4,5,11,14,2,7,9,16$$has the property that is needed for the problem. Thus we have our sequence for $d=1$. By our scaling freedom, we scale by any square to get that $d=t^2$, where $t$ is an arbitrary integer.

k.vasilev wrote: Determine all positive integers $d,$ such that there exists an integer $k\geq 3,$ such that One can arrange the numbers $d,2d,\ldots,kd$ in a row, such that the sum of every two consecutive of them is a perfect square. I claim that the answer is all perfect squares. First, I will prove that any $d$ which satisfies the condition must be a perfect square. Then, I will prove that every perfect square is a solution. Let the numbers $d,2d,\dots , kd$ be rearranged as $a_1,a_2,\dots,a_k$ in a row, and let $b_j=\frac{a_j}{d}$ for every $1\leq j\leq k$. Now, let $d=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_n^{\alpha_n}$ be the decomposition of $d$ in primes. Let's suppose $d$ isn't a perfect square. Then, WLOG, $\alpha_1$ is odd. Then, it is satisfied $$ a_i+a_{i+1}\equiv 0\mod p_1^2 \thickspace \Longrightarrow \thickspace b_i+b_{i+1}\equiv 0\mod p_1 \thickspace \forall \thickspace 1\leq i\leq k-1 $$That is, $$ b_1+b_2\equiv b_2+b_3\equiv \dots \equiv b_{i-1}+b_i\equiv 0\mod p_1 $$Hence, $$ \left\{\begin{array}{lll} b_1\equiv b_3\equiv \dots \mod p_1 \\ b_2\equiv b_4\equiv \dots \mod p_1 \\ \end{array} \right. $$This implies that the elements of the set $\{b_1,b_2,\dots ,b_k\}=\{1,2,\dots, k\}$ have only two different congruences modulo $p_1$. Since $k\geq 3$, then the only possible value for $p_1$ is $p_1=2$. In other words, the only possible prime $p$ such that $v_p(d)$ is odd is $p=2$. However, I claim that $v_2(d)$ must be even. The set $\{1,2,\dots,k\}$ contains at least an even and an odd element. Therefore, the permutation $b_1,b_2,\dots,b_k$ contains at least a pair of consecutive numbers $(b_i,b_{i+1})$ such that one of them is odd and the other one is even, implying $b_i+b_{i+1}\equiv 1\mod 2$. Let's suppose $v_2(d)$ was odd. Then, $$ b_1+b_2\equiv b_2+b_3\equiv \dots \equiv b_{k-1}+b_k\equiv 0\mod 2 $$We have reached a contradiction, and hence we have proved that there doesn't exist any prime $p$ such that $v_p(d)$ is odd. In other words, $d$ must be a perfect square. Now, let's prove that any perfect square is a solution. Let $d=m^2$ be a perfect square. Then, there exists an integer $k=15$ such that the numbers $d,2d,\dots, 15d$ can be arranged in a row, such that the sum of any two consecutive numbers is a perfect square. In particular, the numbers can be arranged as follows: $$ 8d, \, d, \, 15d, \, 10d, \, 6d, \, 3d, \, 13d, \, 12d, \, 4d, \, 5d, \, 11d, \, 14d, \, 2d, \, 7d, \, 9d $$It is easy to check that such arrangement works, and so the answer is all perfect squares.

This strongly reminded me of 24BSLC2. I don't think they're very similar but having done that problem strongly motivated the solution, especially the construction for $n=1$ here. The answer is all integers $d$ such that $d$ is a perfect square. We say an integer $d$ is good if it satisfies the desired conditions. It is not hard to see that an integer $d$ is good if and only if $p^2d$ is good, for any prime $p$ since the sum of any pair of consecutive terms of a sequence of the given form for $p^2d$ is divisible by $p^2$. Thus, in order for the sum to be a square, the sum of the terms, divided by $p^2$ must also be a square. Now, we simply need to check which $d$ are good for square-free $d$. If $d=1$, it is in fact good. To see why consider the following construction. \[8,1,15,10,6,3,13,12,4,5,11,14,2,7,9\]Thus, all square $d$ are also good by the previous observation. Now, if $d$ is not 1, and is square free, it is a product of distinct primes, i.e $d=p_1p_2\dots p_k$. Now, say there exists a term in the sequence $d,2d,\dots, kd$ which is divisible by $p_1^2$. If it is ever a neighbor of a term which is not divisible by $p_1^2$ (but divisible by $p_1$ clearly), then their sum is divisible by$p_1$ but not $p_1^2$, so it is not a perfect square. Thus, all such terms can only be neighboring to terms divisible by $p_1^2$. But clearly, there exists terms which are not divisible by $p_1^2$ in this sequence, so if there exists a term divisible by $p_1^2$ we have a clear contradiction. Thus, the sequence cannot include a number divisible by $p_1^2$. But note that when $d=p_1p_2\dots p_k$, then its neighboring term is atleast $(p_1p_2\dots p_k)^2-p_1p_2\dots p_k$ since $(p_1p_2\dots p_k)^2$ is the smallest square which is a multiple of $p_1p_2\dots p_k$. If it is greater than $(p_1p_2\dots p_k)^2 - p_1p_2\dots p_k$, then we immediately have that $(p_1p_2\dots p_k)^2$ is in this sequence, which is divisible by $p_1^2$. So, the only possible neighbor for $p_1p_2\dots p_k$ is $(p_1p_2\dots p_k)^2-p_1p_2\dots p_k$. But, note that since one of $p_1p_2\dots p_k$ and $(p_1p_2\dots p_k)^2-p_1p_2\dots p_k$ must have some other neighbor unless $p_1p_2\dots p_k-1=2$, there must exist some term which is atleast \[(2p_1p_2\dots p_k)^2-p_1p_2\dots p_k \geq (p_1p_2\dots p_k)^2\](if it is a neighbour of 1) or \[(2p_1p_2\dots p_k)^2 - (p_1p_2\dots p_k)^2 + p_1p_2\dots p_k > (p_1p_2\dots p_k)^2\](if it is a neighbour of $(p_1p_2\dots p_k)^2-p_1p_2\dots p_k$) which implies that $(p_1p_2\dots p_k)^2$ is in this sequence anyways, so we must have $p_1p_2\dots p_k-1=2$ (then there exists no multiples of $p_1p_2\dots p_k$ between $p_1p_2\dots p_k$ and $(p_1p_2\dots p_k)^2-p_1p_2\dots p_k$) but this implies that the sequence has at most 2 elements, which is again a clear contradiction since we are given that $k\geq 3$. Thus, all the answers are as claimed and we are done.