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 nn such that:

  • n1(mod2)n\equiv 1 (\mod 2)

  • You can return to the same block in the infinite chess board using nn knight moves.

If you find any answer then please comment

Note by Zakir Husain
1 month, 1 week ago

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1 vote

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

Here is the proof

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


Let number of moves of type mim_i be nin_i

Now, as total moves are nn, so

i=18ni=n\sum_{i = 1}^{8} n_i = n

Also, for knight to return to its initial position

2n1+2n2+n3+n8=2n5+2n6+n7+n4(1)2n_1 + 2n_2 + n_3 + n_8 = 2n_5 + 2n_6 + n_7 + n_4 \ldots\ldots (1) 2n3+2n4+n2+n5=2n7+2n8+n1+n6(2)2n_3 + 2n_4 + n_2 + n_5 = 2n_7 + 2n_8 + n_1 + n_6 \ldots\ldots (2)

Now, as nn is odd, following conditions arrive

  • One among nin_i is odd and other are even.

  • Three among nin_i are odd and other are even.

  • Five among nin_i are odd and other are even.

  • Seven among nin_i are odd and other are even.

Now, as we can see, in (1)(1) every term with coefficient 11 has coefficient 22 in (2)(2) and vice-versa.

Condition 11

Due to symmetry, considering any of them to be odd will work. Let n1n_1 be odd.

If n1n_1 is odd, then equation (2)(2) will be dissatisfied. So, condition 11 won't work.

Condition 22

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

  • All three odd terms are of coefficient 22 in equation (1)(1), then all three will be of coefficient 11 in equation (2)(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 22 and others are of coefficient 11. Then also, applying above logic, equation (2)(2) won't be satisfied

So, Condition 22 won't work

Conditions 33 and 44

Similar to logic of condition 22 we can prove that conditions 33 and 44 won't work.


So, there exists no odd nn with above conditions.

Hope it helps. :)

Aryan Sanghi - 1 month, 1 week ago

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Great proof.

Zakir Husain - 1 month, 1 week ago

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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 n1(mod 2)n \equiv 1 (\bmod~ 2) is equivalent to saying nn is odd. And for anyone interested, "\bmod~" gets rid of the annoying space before the "mod". :)

David Stiff - 1 month, 1 week ago

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Very elegant and concise solution. :) @David Stiff

Aryan Sanghi - 1 month, 1 week ago

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Thank you!

David Stiff - 1 month ago

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Very beautiful proof, thanks for that.

Zakir Husain - 1 month, 1 week ago

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Your welcome! Glad you liked it.

David Stiff - 1 month ago

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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.

Mahdi Raza - 1 month ago

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Thank you.

David Stiff - 4 weeks, 1 day ago

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All I know is since that 1(mod2)=1,n=11 (\mod 2) = 1, n = 1

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

Yajat Shamji - 1 month, 1 week ago

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