The Mystic Circle

The compulsory links -

$4-32-17$

$5-31-18$

$6-30-19$

$7-29-20$

$8-28-21$

$9-27-22$

$10-26-23$

$11-25-24$

Now $18$ can only be linked with $7$ , So we have $5-31-18-7-29-20$ . And $20$ can only be paired with $16$ which can only be paired with $9$ so we have $5-31-18-7-29-20-16-9-27-22$

And $19$ can only be paired with with $17$ so we have $4-32-17-19-30-6$

Beyond this point is case work along with logic.

$5$ can be paired with either $4$ or $11$ .

$22$ can be paired with $3$ or $14$

$4$ can be paired with $5$ , $12$ , or $21$ .

$6$ can be paired with $3$ or $10$ .

Trying out all options, leads to the one unique solution above.

The number of times the squares $4, 9, 16, 25, 36, 49$ appear are $0, 2, 4, 6, 12, 8$ respectively. The answer is $0 + 4 +12 + 24 + 60 +48 =148$ .

Love this problem! Thank you for sharing! :)

I propose a solution in python 3.4:

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def possibleneighbours(n):
    res=[]
    for i in range(2,8):
        m = i*i - n
        if m > 0 and m <= 32 and m != n:
            res.append(m)
    return res

allsolutions = []

def browse(mypath):
    global allsolutions
    currentnode = mypath[-1]
    neighbours = possibleneighbours(currentnode)
    if len(mypath)== 32 and mypath[0] in neighbours:
        allsolutions.append(list(mypath))
    disp = list(set(neighbours) - set(mypath))
    for nextnode in disp:
        mypath.append(nextnode)
        browse(mypath)
        mypath.pop()


def computeresult():
    global allsolutions
    allsolutions = []
    browse([32])
    path = allsolutions[0]
    tot = 0
    for i in range(32):
        j = (i + 1)%32
        tot += ((path[i] +  path[j])**.5) -1
    return tot  

computeresult()

Because there are eight compulsory chains formed by three numbers each to begin with , in which the central number decreases from $32$ to $25$ (leaving exactly the eight lonely numbers $1,2,3,12,13,14,15,16$ ), and each of these chains has two numbers that sum up to $49$ and two numbers that sum up to $36$ , we have, right at the beginning, the following conditions:

$F\geq8,\quad E\geq8.$

But it is impossible to obtain another $49$ by adding two numbers from different chains or by adding a tip number of any chain and any lonely number . Therefore, $F=8$ .

But we can also observe that because of the compulsory links $(17-19),\space(18-7),\space(20-16-9),\space(1-8),\space(15-10),\space(23-2-14)$ and $(3-13-12)$ , there are six compulsory chains to begin with, which are:

$\begin{aligned} & (4-32-17-19-30-6) \\ & (5-31-18-7-29-20-16-9-27-22) \\ & (1-8-28-21) \\ & (15-10-26-23-2-14) \\ & (3-13-12) \\ & (21-25-24) \end{aligned}$

This means we have exactly one $9$ , two $16$ , five $25$ and two $36$ more, so, even before starting the "guess and check", we have the following conditions:

$F=8,\quad E\geq10\,\quad D\geq5,\quad C\geq2,\quad B\geq1.$

Therefore, we can simplify the structural equation of the Mystic Circle from

$4A+9B+16C+25D+36E+49F=2\sum_{n=1}^{32}n=1056$

to

$4A+9\beta+16\gamma+25\delta+36\varepsilon=1056-8\times49-10\times36-5\times25-2\times16-1\times9=138$

where $B=1+\beta,\space C=2+\gamma,\space D=5+\delta,\space E=10+\varepsilon$ and $A+\beta+\gamma+\delta+\varepsilon=6$ .

From the last condition, because we are adding five non-negative integers to make a six , by the pigeonhole principle we know that at least one of these integers must be at least equal to two .

At this point, note that $A\leq2$ , because if $A=3$ , for example, even a value of $\varepsilon=3$ wouldn't be enough for the total to be $138$ , because $3\times4+3\times36=120<138$ .

Because $138\equiv0\pmod{23}$ , applying some modulo- $23$ arithmetic to the remainders $4,9,16,2,13$ of the squares $4,9,16,25,36$ when divided by $23$ , it can be shown that $A<2$ , because the only solution with $A=2$ , which is $A=2,\gamma=3,\varepsilon=1$ , is not valid since $2\times4+3\times16+1\times36=92\neq138$ .

By the same logic, it can be shown that $A\neq1$ , because the only solution with $A=1$ , which is $A=1,\varepsilon=5$ , is not valid since $1\times4+5\times36=184\neq138$ . Therefore, $A=0$ .

Knowing this, and applying the same modulo- $23$ arithmetic, it can be shown that the only valid solution is $\beta=1,\gamma=2,\delta=1,\varepsilon=2$ , which corresponds to

$A=0,B=2,C=4,D=6,E=12,F=8$

and yields the correct answer:

$1\times0+2\times2+3\times4+4\times6+5\times12+6\times8=\boxed{148}.$

Therefore, the solution of the Mystic Circle is unique.

Python solution:

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NUMBERS = range(1, 33)
SQUARES = (4, 9, 16, 25, 36, 49)

def main():
    streak = []
    nums = list(NUMBERS)
    n = nums.pop(0)
    streak.append(n)
    success = recurse_add(n, nums, streak)

    if success:
        print '******\nStreak\n******'
        print streak

    squares = [(streak[i] + streak[i-1]) for i in range(0, len(streak))]
    print '******\nSquares\n******'
    print squares

    sq_count = []
    for s in SQUARES:
        sq_count.append(squares.count(s))
    print '******\nSquare count\n******'
    print sq_count

    answer = 0
    for m in range(1, 7):
        answer += m*sq_count[m-1]

    print '******\nAnswer\n******'
    print answer

def recurse_add(n, nums, streak):
    if not nums:
        return (streak[0] + streak[-1]) in SQUARES
    added = False
    for i in range(0, len(SQUARES)):
        a = SQUARES[i] - n
        if a < 1:
            continue
        if a in nums:
            nums.remove(a)
            streak.append(a)
            added = recurse_add(a, nums, streak)
            if not added:
                streak.remove(a)
                nums.append(a)
    return added 

Output:

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******
Streak
******
[1, 8, 28, 21, 4, 32, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 5, 31, 18, 7, 29, 20, 16, 9, 27, 22, 14, 2, 23, 26, 10, 15]
******
Squares
******
[16, 9, 36, 49, 25, 36, 49, 36, 49, 36, 9, 16, 25, 36, 49, 36, 16, 36, 49, 25, 36, 49, 36, 25, 36, 49, 36, 16, 25, 49, 36, 25]
******
Square count
******
[0, 2, 4, 6, 12, 8]
******
Answer
******
148 

Going with compulsory pairs (numbers that could only form squares in two ways, either directly or after some numbers had been eliminated) I got to part rings 1-8-28-21 + 5-31-18-7-29-20-16-9-27-22 + 6-30-19-17-32-4 + 10-26-23-2-14 + 11-25-24 + 3-13-12. At this point I was stuck a bit, but 15 gave me the next step: It can be coupled with 1, 10 and 21, but 1 and 21 are at 2 ends of the same part ring, so we cannot couple it with both of these, which means it must be coupled with 10. After this, the numbers all fell into 3 part-rings, from which the full ring was easily created.