Let the unknown integer by n.
2n = 6 + 2n2 : 2n2 - 2n + 6 = 0
For a quadratic equation an2 + bn + c = 0 then the equation has no real roots if
b2 - 4ac < 0 : In the example b2 - 4ac = 4 -48 = -44 as this < 0 then the original expression cannot have a real integer solution.
If a and b are integers, then a times b is an integer.
The sum of three consecutive integers is -72
8
The smaller integer is 6, the larger integer is 32
10-11-12
9,11,13
If a and b are integers, then a times b is an integer.
The sum of three consecutive integers is -72
That has no integer solution. Three times an integer is another integer; if you subtract to integers, you get an integer again, not a fraction.
8
It is two times the magnitude of the integer.
The smaller integer is 6, the larger integer is 32
-10
Let's denote the unknown integer as "x". So now we have two integers, "x" and "4x" because one integer is 4 times the other. So the sum of x+4x= 5x 5x = 5 So x=1
The integers are 5 and 7.
The other integer is -18 because -5 times -18 = 90
The integers would begin with 10.