11.4 Another participant in the meaningless activity contest brought with him $N$ unit squares, from which he immediately, in front of the amazed jury, formed a rectangle with sides differing by 9. Not stopping there, the participant then formed a large square from these same $N$ squares, but this time 6 squares were left over. Find the possible values of $N$.
Solution 1: Let $x$ be the smaller side of the rectangle, and $y$ be the side of the square. Then $N=x(x+9)=y^{2}+6$. Multiplying both sides of the equation by 4, we get $2 x(2 x+18)=$ $(x+9)^{2}-81=4 y^{2}+24,(2 x-2 y+9)(2 x+2 y+9)=105$. In this product, the factor $2 x+2 y+9$ is positive and no less than 13, and the second factor is an integer. The divisors of the number 105, greater than 12, are $15,21,35$ and 105 (then $2 x-2 y+9$ is respectively equal to $7,5,3$ and 1). Solving the system of linear equations in each case, we arrive at the solutions $(1,2),(2,4),(5,8)$ and $(22,26)$, which correspond to the values of $N=10,22,70,682$ respectively.

Solution 2: The equation $x(x+9)=y^{2}+6$ can be solved differently. On the one hand, $9 x>6$, so $y^{2}+6=x^{2}+9 x>x^{2}+6, y>x$. On the other hand, $y^{2}+6=x^{2}+9 x<x^{2}+10 x+31=$ $(x+5)^{2}+6, y<x+5$. Thus, the value of $y-x$ can be $1,2,3$ or 4. In each of the four cases, the equation reduces to a linear one, the solution of which allows finding one of the four values of $N$.

## Criteria:

- When listing divisors, only natural divisors are considered, and the positivity of the two factors is not obvious - up to 2 points are deducted;
- Only the answer - 0 points.