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We call a coloring of a square with black and white colors nice if none of black square share same segments.Prove that the nice coloring cases of a $10\times 10$ square is more than $10^{15}$ and less than $10^{25}$.

It is to hard to handle with calculating it because we should divide every case to $2$ case if the $1\times 1$ square is black or white then how can we make the prove?I have tried using induction but it doesn't work for $1\times 1$ case so I didn't go anymore.Any hints?

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    "if non of black square doesn't have same segments." Did not get this line. What is a segment ? – 2017-01-06
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    @A---B Sth that connects two vertexes. – 2017-01-06
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    @TahaAkbari Still don't get your problem. You mean you have a $10 \times 10$ checkboard (with nonstandard coloring) so coloured, that no two black squares are next to each other, i.e. share a side? – 2017-01-06
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    @ctst Yes then we want to know how many boareds are like that. – 2017-01-06

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The total number of ways to color each $2\times2$ square is $1+4+2=7$. So the total number of nice coloring is less than $7^{25}$. So that settles the upper inequality.

For the lower inequality consider the chessboard and consider all of the $2^{50}$ colorings that result from repainting some of the black squares white. We have $2^{50}>10^{15}$ so we are done.

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    if you want to prove $2^{50}>10^{15}$ you can notice $2^{10}> 10^3$ and so $2^{50}>10^{15}$ – 2017-01-06