$y = \frac{x^2 - 6x + 1}{x - 3} = \frac{x^2 - 6x + 9 - 8}{x - 3} = \frac{(x - 3)^2 - 8}{x - 3} = x - 3 - \frac{8}{x - 3}$
Дробь $\frac{8}{x - 3}$ будет целой при x − 1 = {±1; ±2; ±4; ±8}.
x − 3 = −1
x = 2
$y = 2 - 3 - \frac{8}{2 - 3} = -1 - \frac{8}{-1} = -1 + 8 = 7$
 
x − 3 = 1
x = 4
$y = 4 - 3 - \frac{8}{4 - 3} = 1 - \frac{8}{1} = 1 - 8 = -7$
 
x − 3 = −2
x = 1
$y = 1 - 3 - \frac{8}{1 - 3} = -2 - \frac{8}{-2} = -2 + 4 = 2$
 
x − 3 = 2
x = 5
$y = 5 - 3 - \frac{8}{5 - 3} = 2 - \frac{8}{2} = 2 - 4 = -2$
 
x − 3 = −4
x = −1
$y = -1 - 3 - \frac{8}{-1 - 3} = -4 - \frac{8}{-4} = -4 + 2 = -2$
 
x − 3 = 4
x = 7
$y = 7 - 3 - \frac{8}{7 - 3} = 4 - \frac{8}{4} = 4 - 2 = 2$
 
x − 3 = −8
x = −5
$y = -5 - 3 - \frac{8}{-5 - 3} = -8 - \frac{8}{-8} = -8 + 1 = -7$
 
x − 3 = 8
x = 11
$y = 11 - 3 - \frac{8}{11 - 3} = 8 - \frac{8}{8} = 8 - 1 = 7$
Ответ: {(−5;−7),(−1;−2),(1;2),(2;7),(4;−7),(5;−2),(7;2),(11;7)}