# A new chess problem!

Suppose you have an infinite chess board and a knight $♘$ in some block.

Now you need to find at least one natural number $n$ such that:

Note by Zakir Husain
1 month, 1 week ago

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@Zakir Husain Sir, there will exist no odd $n$ with the above conditions

Here is the proof

First, let us number the moves possible by night in an anticlockwise manner

 $m_6$ $m_5$ $m_7$ $m_4$ ♘ $m_8$ $m_3$ $m_1$ $m_2$

Let number of moves of type $m_i$ be $n_i$

Now, as total moves are $n$, so

$\sum_{i = 1}^{8} n_i = n$

$2n_1 + 2n_2 + n_3 + n_8 = 2n_5 + 2n_6 + n_7 + n_4 \ldots\ldots (1)$ $2n_3 + 2n_4 + n_2 + n_5 = 2n_7 + 2n_8 + n_1 + n_6 \ldots\ldots (2)$

Now, as $n$ is odd, following conditions arrive

• One among $n_i$ is odd and other are even.

• Three among $n_i$ are odd and other are even.

• Five among $n_i$ are odd and other are even.

• Seven among $n_i$ are odd and other are even.

Now, as we can see, in $(1)$ every term with coefficient $1$ has coefficient $2$ in $(2)$ and vice-versa.

# Condition $1$

Due to symmetry, considering any of them to be odd will work. Let $n_1$ be odd.

If $n_1$ is odd, then equation $(2)$ will be dissatisfied. So, condition $1$ won't work.

# Condition $2$

So, if equation $(1)$ is satisfied, then following are possible

• All three odd terms are of coefficient $2$ in equation $(1)$, then all three will be of coefficient $1$ in equation $(2)$. As there will be three odd terms, so both LHS and RHS have to be of different parity.

• One odd term is of coefficient $2$ and others are of coefficient $1$. Then also, applying above logic, equation $(2)$ won't be satisfied

So, Condition $2$ won't work

# Conditions $3$ and $4$

Similar to logic of condition $2$ we can prove that conditions $3$ and $4$ won't work.

# Conclusion

So, there exists no odd $n$ with above conditions.

Hope it helps. :)

- 1 month, 1 week ago

Great proof.

- 1 month, 1 week ago

If we assume the infinite chessboard is coloured like a regular chessboard, with alternating black and white squares, we can show that every time the knight moves, it will go from a black square to a white square, or vice-versa. This is because we move 2 squares in one direction, landing on our original colour, and then 1 square in another direction, landing on the opposite colour.

Let's assume we start on a black square. Then, if we make an odd number of moves, we know we must land on a white square (from black to white for 1 move, black to white to black to white for 3 moves, and so on). Thus, it will be impossible to return to our black square in an odd number of moves. The same holds for if we started with a white square.

Thanks @Aryan Sanghi for pointing out that $n \equiv 1 (\bmod~ 2)$ is equivalent to saying $n$ is odd. And for anyone interested, "\bmod~" gets rid of the annoying space before the "mod". :)

- 1 month, 1 week ago

Very elegant and concise solution. :) @David Stiff

- 1 month, 1 week ago

Thank you!

- 1 month ago

Very beautiful proof, thanks for that.

- 1 month, 1 week ago

- 1 month ago

Every move, in any direction you take, will lead to a different color. Hence to return to the same color, you must require an even number of moves, and an odd number of moves won't satisfy.

Well observed and written.

- 1 month ago

Thank you.

- 4 weeks, 1 day ago

- 1 month, 1 week ago

Interesting...

All I know is since that $1 (\mod 2) = 1, n = 1$

Don't blame me of my lack of modulo knowledge, yes?

- 1 month, 1 week ago