# Find a way out in $8 \times 8$ square

Here is one of the mathematic contest problem that my teacher has given me. He said me to solve or find the reason why it cannot be solved. But I was not able to do it. And my searches for similar answers was not effective.

Given: $8 \times 8$ square which consists of 64 squares.

Objective: Draw connected lines from right-bottom square to top-left square which will include EVERY square.

Rules: You must start from right-bottom square and finish at top-left square (See the link). You can only draw a line up, down, left, right directions. Moving diagonally and crossing lines is not allowed.

Sorry for bad English. Please warn me, if you misunderstood the problem. As an example, I provide the solved $7 \times 7$ square with the rules above.

The example $7 \times 7$ square

The $8 \times 8$ square

Imagine this is a chess board, i.e. it's colored in black and white alternatively. Now note that when you make a move you always go from white to black or from black to white. Say the bottom-right corner is black. Then the top-left corner is also black. But if your line goes trough every square exactly once, then it's 64 square long (which is even), which means that the last square should be white (because they alternate). This is a contradiction, which proves that such a line does not exist.

There is no solution. The number of moves is $63$, so the start- and endplace must have different colors, which is not the case for opposite corners.

Imagine you are moving rook from a black spot 1 square per move . So on 63rd move our rook needs to be on white spot which is last rowth or last columnth. But if we place a rook on right bottom black square only after even number of moves it will again be on same color abiding by above mentioned condition. But $63$ is odd so its contradictory so such thing will never happen.