## Two Proofs for a Five Point Star

```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    ?
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.
```