Suppose we arrange the numbers 1-10 on a 5-point star, putting a number where any two lines meet. Each line then has four numbers on it. Can we arrange the numbers so that each line adds up to 22? This is an interesting problem. One way of approaching it might be to consider the parity requirements imposed by making each line sum to an even number (22). It follows that each line must have exactly two odd numbers and two even numbers from the range 1 to 10, and there are essentially only two ways of doing this, up to rotations, reflections, and complements 0 1 0 1 1 0 1 0 0 1 1 1 1 1 1 0 0 0 0 0 So, given the five equations for the row sums along with the equation for the sum of the squares of the 10 numbers A^2 + B^2 + C^2 + D^2 + E^2 + F^2 + G^2 + H^2 + I^2 + J^2 = 385 we can eliminate all fice of the odd numbers (or all five of the even numbers) to give an equation of the form 605 - 22 (A+B+E+I+J) + 4 (A^2 + B^2 + E^2 + I^2 + J^2) - 4 (AB+BI+IJ+JE+EA) + 2 (AJ+EI+JB+IA+BE) = 385 (We get this same equation, with different labels, regardless of which parity pattern we choose, and whether we eliminate the odds or the evens.) By symmetry, this equation must be satisfied if we set A,B,E,I,J equal to the five even numbers 2,4,6,8,10 AND if we set them equal to the five odd numbers 1,3,5,7,9 in some order. Let's take the odd numbers. Happily the first two parenthetical quantities are symmetrical in the five numbers, so we know (A+B+E+I+J) = 25 (A^2 + B^2 + E^2 + I^2 + J^2) = 165 The next two terms are not entirely symmetrical, but we can improve it a little by adding the sum of every product of two numbers to each side. This quantity is (AE+AJ+AI+AB+EJ+EI+EB+JI+JB+IB) = 230 so the equation becomes 100 = 3 (AB+BI+IJ+JE+EA-AJ-EI-JB-IA-BE) which has no integer solutions because 100 is not a multiple of 3. This completes the proof. ========= Since this proof evidently relies on mod 2 (parity) and mod 3 considerations, it seems natural to think there might be a more direct proof from consideration of the possibilities modulo 6. The ten numbers we need to distribute are 1 2 3 4 5 0 1 2 3 4 (mod 6) where odd/even parity is preserved, so each line in the pentagram must contain exactly two elements from the multiset {1,3,5,1,3} and two from the multiset {2,4,0,2,4}. Also, each line must sum to 22=4 (mod 6). First, note that each of the five lines intersects with each of the other lines in exactly one node, and each node is the intersection of exactly two lines. Now consider the two lines that meet at the single residue 5, and consider the possible ways of filling out the rest of those two lines. The sum of positive residues must be 4 or 10 or 16, which are the only values congruent to 4 (mod 6) reachable by a sum of four of these positive residues. Thus, the remaining three numbers in each of the two lines must sum to 1, 5, or 11. Also, one of the three numbers must be odd and the other two must be even. The only sets of three (remaining) elements with these properties are a={4,4,3} b={4,1,0} c={3,2,0} d={2,2,1} (Note that {3,1,1} is disqualified on odd/even parity, and there is no {1,0,0} because we have only one "0".) Each of the lines intersecting at the number "5" must consist of one of these multisets. We can review the possible combinations as follows aa Impossible, because it requires four 4's ab Impossible, because it requires three 4's ac ? ad ? bb Impossible because it requires two 0's bc Impossible becasue it requires two 0's bd ? cc Impossible because it requires two 0's cd Impossible because it requires three 2's dd Impossible because it requires four 2's So this reduces the possibilities to just three. But consider the case ac. In this case the two lines that meet at "5" consist of a={4,4,3} and c={3,2,0}, which leaves only the residues {2,1,1} to fill in the remaining three location. But this is impossible, because wherever the 4's are placed in their line they each lie on one other line that intersects the c line once (at a node other than where the 5 is located), and then contains two elements from the set {2,1,1}. Neither of the 4's can insersect with either 0 or 2 in the c line, because then no two elements of {2,1,1} will make the sum congruent to 4 (mod 6). Thus combination ac is impossible. The case ad has the two lines that meet at "5" consisting of a={4,4,3} and d={2,2,1} which leaves only the set {3,1,0} to fill in the remaining three locations. Again each 4 on the "a" line is also on one other line that intersects with the d line and then contains two elements from {3,1,0}. Clearly neither of the 4's can intersect with 1 in the d line, because then no two elements of {3,1,0} will make the sum congruent to 4 (mod 6). Thus, each 4 must be on a line that intersects with a 2 on the d line. However, this means that they both need {3,1} from the remaining numbers, which is impossible. (They could share the 3, but then one of them would have to take the 0.) Therefore, the combination ad is impossible. This leaves only the case bd, where b={4,1,0} and c={2,2,1}. The remaining three elements are then {4,3,3}. This combination is also impossible, because the "4" in line b also lies in a line that intersects line c at one point and then is filled out with two elements of {4,3,3}. If the 4 in b intersects with the 1 in c, then we need two elements of {4,3,3} that sum to 5 or 11, but the only sums of two elements of this set are 6 and 7. On the other hand, if the 4 in b intersects with 2 in c, then we need two elements of {4,3,3} that sum to 4 or 10, which again we don't have. Therefore, combination bd is impossible, completing the proof.

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