Let consider a square $10$x$10$ and write in the every unit square the numbers from $1$ to $100$ Let   consider   a  square $10\times 10$   and  write  in  the   every   unit  square    the   numbers   from  $1$  to  $100$   such  that   every  two   consecutive  numbers  are  in   squares   which  has  a  common edge.  Then  there  are  two  perfect  squares   on  the  same line  or  column. Can  you   give  me  an  hint? How  to  start?
 A: We note the following: 


*

*Write the coordinates of $k$ as $(i_k,j_k)$, where $i_k$ is the column that $k$ is in; $i_k \in \{1,2,\ldots, 10\}$; and $j_k$ is the row that $k$ is in; $j_k \in \{1,2,\ldots, 10\}$. Then if $i_k+j_k$ is even, then $i_{k+1} + j_{k+1}$ must be odd, for each $k=1,2,\ldots, 99$.

*If $i_{k^2} + j_{k^2}$ is even, then $i_{(k+1)^2} + j_{(k+1)^2}$ must be odd, as $(k+1)^2-k^2$ is an odd integer, for each $k=1,2,\ldots, 9$.

*We call a square $k^2$ even-even if $i_{k^2}$ and $j_{k^2}$ are both even. 
and we call a square $k^2$ odd-odd if $i_{k^2}$ and $j_{k^2}$ are both odd. We call a square mixed otherwise. Then if $k^2$ is odd-odd or even-even, then $(k+1)^2$ must be mixed.
So from 3 we have the following:

 4. Precisely 5 squares are mixed and precisely 5 squares that are either even-even or odd-odd. 

But this is impossible unless a row or column has at least 2 squares:

 Indeed: Either at least 3 of the squares $k^2; k=1,2,\ldots, 10$; are even-even, or at least 3 of the squares are odd-odd. LEt us assume that 3 of the squares are even-even. Then if every row and column has exactly one square, then of the 5 mixed squares, only 2 can be in an even column (as 3 of the even columns were already taken by the 3 even-evens and so there are only 2 even columns left). And likewise, only 2 can be in an even row. But this implies that at least one (i.e. $5-2-2$) of the 5 mixed squares is odd-odd after all, which contradicts 4. above. [The likewise holds by the same line of reasoning holds if 3 of the squares are odd-odd.]

A: Here are several successively more revealing hints, hidden behind spoilers in case you want to try the problem after only reading one or two.
Hint 1: 
Color your $10\times 10$ board like a checkerboard. What can you say about the colors of the squares containing perfect squares?
More specifically:

 How does the color of square $1$ compare to that of $4$? And how does $4$ compare to that of $9$? Etc. 

Hint 2: 

 In general, show that for any $10$ squares in pairwise different rows and columns, an even number of these squares must be black.

  Assuming that a path where the perfect squares are in different rows and columns exists, combine this fact with the conclusion of Hint $1$ to get a contradiction.

Hint 3:

 This goes into more detail about how to prove the first sentence of Hint $2$.

 Suppose there are $10$ squares in pairwise different rows and columns. A square in row $i$ and column $j$ is black if and only if $i+j$ is even.

 Suppose square in row $i$ is in column $\pi_i$, where $\pi$ is a permutation of $\{1,2,\dots,n\}$. Then the summation $\sum_{i=1}^{10}(i+\pi_i)$ is equal in parity to the number of black squares, so you need to prove this summation is even. 

