I have to arrange the numbers $1,\ 2,\ ...,\ 9$ on the points in the lines such that every two lines the sum of their elements is equal.
My Solution:
We already know that $1 + 2 + ... + 9 = 45$
Let $x_1,\ x_2$ and $x_3$ the numbers in the fist line (from Top-Down)
$x_3,\ x_4$ and $x_5$ the numbers in the second line
$x_5,\ x_6$ and $x_7$ the numbers in the third line
and $x_7,\ x_8$ and $x_9$ the numbers in the fourth line
Then $$45 + x_3 + x_5 + x_7 = 4k$$
The smallest multiple of $4$ this is grether than 45 is $48$ That mean $$x_3 + x_5 + x_7 = 3$$ well the equation is unsolveable for 3. Let go head to the next multiple of 4 that is 52 that means $$x_3 + x_5 + x_7 = 7$$ The equation is solveablefor $$x_3 = 1 \text{ and } x_5 = 4 \text{ and } x_7 = 2$$ We can immediately $x_4$ because $x_3 + x_4 + x_5 = 13$ Hence $x_4 = 13 - 7 = 6$
Further $x_1 = 3,\ x_2=9,\ x_3=1,\ x_4=8,\ x_5=4,\ x_6=7,\ x_7=2,\ x_8=6\text{ and } x_9=5$
Considerung the solution is came across an other solution. Well my question is there a way to cover the solution space.