# Two sums on a chessboard

There is an $n\times n$ chessboard with an even positive integer $n$. We put all the numbers $1,2,...,n^2$ into the squares of the chessboard, where each number appears once and only once and each square has one and only one number. The sum of the numbers put in the black squares is $B$ and the sum of the numbers put in the white squares is $W$.

Find all $n$ so that we can achieve $\frac{B}{W}=\frac{39}{64}$.

• Not enough information. What order do the numbers go on the board. Can we arrange them as we like? – InternetHobo Mar 19 '16 at 16:30

I believe there has to be more than one answer to this question; For example, there is a solution for $n=206, ...,103*2m$ where $m>0$ and integer.

So there are actually infinite solutions since $B+W=103x$ where $x$ is an integer, in other words the sum of the values in the $n \times n$ board is the multiple of $103$, there has to be solution with the numbers in the board.

For example, if there is $206 \times 206$ board, the set of numbers becomes $\left \{1,2,3....,206^2 \right \}$. so the sum of these numbers is;

$\frac{206^2*(206^2+1)}{2}$ from the equation $\frac{n*(n+1)}{2}$ fo the sum of a sequence of numbers.

And since this value is multiple of 103, there has to be solution by putting all numbers in this $206 \times 206$ board and squares in it and n is even integer number as well.

• The answer can be all multiples of 206. – Alexis Mar 19 '16 at 16:35
• Also all values of n were requested. It's not multiple solutions if the solution calls for multiple values. – InternetHobo Mar 19 '16 at 16:38
• @Marius why did u take to the power of 2? it should be 1,2,3 and 4. read the question again. the ending has to be $2^2$ – Oray Mar 22 '16 at 16:09
• @Oray. Ignore me. I'm a bit tired. I would be more careful next time. I'll delete my comment to avoid public humiliation. :) – Marius Mar 22 '16 at 16:11
• @Oray. But my theory still stands (kind of). Let's take a 3X3 board. The sum of 1+2+3+4+5+6+7+8+9 = 45. 45 is divisible by 9, but you cannot make any combination so that W/B = 1/8. In this case. B = 8W So W + 8W = 45 This means W = 5. You cannot pick 4 (or 5) numbers to sum to 5. – Marius Mar 22 '16 at 16:19

Partial:

$\frac{B}{W} = \frac{39}{64}$ means
$W = \frac{64*B}{39}$.
Also
$1 +2 + ... + n^2 = \frac{n^2 * (n^2 + 1)}{2}$
this means that $B + \frac{64*B}{39} = \frac{n^2 * (n^2 + 1)}{2}$
Let's make $x = n^2$.
The equation above transforms to
$\frac{103*B}{39} = \frac{x^2 + x}{2}$.
Going further:
$39*x^2 + 39*x-206*B = 0$
Solving this for x I ended up with: \begin{align} x = \frac{-39 \pm \sqrt{39^2 + 4*39*206*B}}{2*39} \\ \end{align}

Since x is an integer this means that $\sqrt{39^2 + 4*39*206*B}$ should be an integer. so $39^2 + 4*39*206*B$ is a perfect square.
In addition, x has to be a perfect square.

I'm stuck after this.