( 1)2x^2-5xy+2y^2+x-2y-6=0 = >(x-2y)(2x-y+ 1)=6

Because both x and y are integers, x-2y and 2x-y+ 1 are also integers. The integer product decomposition of 6 can only be (-6)×(- 1), (-1)×(-6), (-3)×(-2) × (-3), 1×6 and 6× 66.

Solve the equation simultaneously, and abandon the non-integer solution to get the following four groups of solutions: x=- 1, y =1; x=-2，y = 0； x=3，y = 1； x=-2，y=-4

(2)x^4-y^4-20x^2+28y^2= 107 = >(x^2+y^2-24)(x^2-y^2+4)= 1 1

Because both X and Y are integers, X 2+Y 2-24 and X 2-Y 2+4 are also integers. The integer product decomposition of 1 1 can only be (-11) × (-1) and1.

Solve the equation simultaneously, abandon the non-integer solution, and get the following eight groups of solutions: x=-2, y =-3; x=-2，y = 3； x=2，y =-3； x=2，y = 3； x=-4，y =-3； x=-4，y = 3； x=4，y =-3； x=4，y=3

The second question:

Firstly, the possible value of m is determined by remainder analysis.

1) First, m cannot be an even number. Because if m is even, then (5m 2) mod4 = 0 (mod4 stands for the remainder divided by 4), (-6mn) mod4 = 0, 1993mod4 = 1, so (7N2) mod4 = 1, so n is odd.

2) Secondly, m must be divisible by 3. Assuming that m is not divisible by 3, then m=3x+ 1 or m=3x+2, in either case, (5m 2) mod 3 = 2, (-6mn) mod 3 = 0,1993mod 3 =1.

3) According to the conclusions of 1) and 2), we know that m=6x+3.

Secondly, enumeration proves that there is no integer solution.

4) Take 5m 2-6mn+7N2 =1993 as the quadratic equation of n, and get n = [3m+√ (13951-25m 2)]/7 or n = [3m-√ (/kloc-0). Since13951-25m 2 > = 0 means | m | < =23, due to the positive and negative symmetry, we will examine whether m = 3,9,15,21can make (1390/.

The third question:

This topic is actually very simple. We study that the remaining number of red, yellow and blue glass sheets is 3, and the initial state is

& lt0， 1，2 & gt； -& gt； & lt2，0， 1 & gt； -& gt； & lt 1，2，& gt-& gt； & lt0， 1，2 & gt； Cycling to this point, there is no situation that two remainders are 0 at the same time. Painting each glass with the same color means that both remainders are 0, so it is impossible.