Square sum proof

2	9	4
7	5	3
6	1	8

With the given numbers 1,2,3,4,5,6,7,8,9
Is it possible to arrange them in a square given below,
so that the sum either vertically, horizontally, or crossover is the same.
If it is, then how should the numbers be arranged and what must the sum be?

a	b	c
d	e	f
g	h	i

Sum value

The sum of the numbers is 1+2+3+4+5+6+7+8+9 = 45
A sum in the square is based on adding three numbers, and nine numbers are to be used.
This means that the squaresum must be 45 / 3 = 15

Arrangement of the numbers We see the number at place e is to be used four times

a + e + i
b + e + h
c + e + g
d + e + f

center number

To find which number fits in that place we see which numbers can be at that place.
Since the sum must be 15, the sum of the other two numbers without e must be e - 15

The number e is connected to the other numbers a,b,c,d,f,g,h,i, which is the rest of the numbers
The numbers from 1 to 9 is evaluted to see which can fit there.

e=1 ?

then the sum of other numbers must be 15 - 1 = 14

5 + 9
6 + 8

only two sums are valid, there must be four to make it possible

e=2 ?

then the sum of other numbers must be 15 - 2 = 13

4 + 9
5 + 8
6 + 7

only three sums are valid

... Checking all the possibilites for all the numbers = e ?, it turns out only the number e = 5 is valid

e=5 ?

then the sum of other numbers must be 15 - 5 = 10

1 + 9
2 + 8
3 + 7
4 + 6

Four sums are valid, which is required.
The only possible number for e is found

a	b	c
d	5	f
g	h	i

The corner numbers

The numbers a, c, g and i are used three times
and the numbers b, d, f, h are used two times.

In similar way checking the possibilites for a, c, g and i
gives the result where only these four numbers are possible to be put at these places.
These numbers are: 2, 4, 6, 8

e.g.
a + e + i = 15, then a + i = 10 because e = 5
if a = 2 then i is set to 8 because a + i = 10

Since the crossover is set to these numbers 2 and 8, then two other sums which uses this number 2 are required

is a = 2 possible ?

4 + 9
6 + 7

Here are the two other valid sums which give the desired value 13 when 2 is used.

The numbers 2, 4, 6 and 8 is inserted, where the crosssum must be 15

2	b	4
d	5	f
6	h	8

Another method to see what the cornervalues must be,
is to realise that all the cornervalues must be either positive or negative numbers.

They can't be negative because the outer horizontal and vertical lines must also be 15
and there are not enough negative numbers to make this happen.

e.g.
a + b + c must be 15, if a and c is negative, then b must also be negative if the sum must be 15

This figure can be mirrored both vertically or horizontally,
which will give four possible solutions if rest of the numbers can fit the remaining places.
The rest is to see if the rest of the unused numbers can fit. And it's possible

2	9	4
7	5	3
6	1	8

4	9	2
3	5	7
8	1	6

6	1	8
7	5	3
2	9	4

8	1	6
3	5	7
4	9	2

So there are four and only four possible solutions to this figure where the sum must be the same

- epikur

Hosted by www.Geocities.ws