Number puzzle, number arrangement Arrange all of the digits from the numbers 1 through 15 in such a order that the sum of any adjacent pair is a perfect square. No repeats. I can be staring at those number for days. But I wonder if there is a trick.
 A: To perhaps simplify the suggestion I made above: Try writing each number from 1 through 15 in a column.  Next to each, write all the numbers that are greater than that number and that can be added to that number to form a perfect square.  For example:
1: 3, 8, 15
2: 7, 14
And so forth.  For some of these numbers, you'll find your options are limited:  There are very few other numbers they can be paired with.  That will, in turn, restrict where in your string you can put them.  With that in mind, draw a series of fifteen blanks and start putting in the numbers that must go in specific positions.
Let me know if you have questions.
A: The solution is
8 1 15 10 6 3 13 12 4 5 11 14 2 7 9

You could also reverse the order.
The trick is to first realize that the included numbers cover only 4 possible squares - 4, 9, 16, 25.
From there you simply pair numbers to reach these squares.
A: Write down the numbers between $1$ and $15$. Now connect any two numbers iff their sum is a square. Note that you just have to, for each number, calculate its differences from next few bigger squares and connect it to those differences. For example, for $3$, we compute $4-3$, $9-3$, $16-3$, and any larger squares and the differences won't be between $1$ and $15$, so we can stop here.
Now you're looking for a Hamiltonian path through that graph.
A: This answer is based on the original problem statement (reiterated in OP's comment to Anthony's answer), which asks about arranging the digits from the numbers $1$ to $15$, so that the sum of any adjacent pair is a perfect square.

Notice that there are eight "$1$"s available, from $1$, $10$, $11$, $12$, $13$, $14$, $15$. So, the list looks like this:
$$\dots 1 \dots 1 \dots 1 \dots 1 \dots 1 \dots 1 \dots 1 \dots 1 \dots$$
The leading and trailing "$\dots$"s may be empty, but since $1+1$ is not a perfect square, each pair of "$1$"s must be separated by at least one other digit. 
The only digits that can go next to a "$1$" are: "$0$" (of which you have one available, from $10$), "$3$" (of which you have two, from $3$ and $13$), and "$8$" (of which you have one, from $8$). That's only four available candidates, but we have at least seven $1$-separating slots to fill.
The puzzle, as originally stated, has no solution.
