Multiple Choice Question (MCQ)
How many integer solution sets (x,y) for the equation x^3+6x^2+5x=y^3y+2?

✓
0

×
1

×
3

×
infinite
How many integer solution sets (x,y) for the equation x^3+6x^2+5x=y^3y+2?
0
1
3
infinite
x^3+6x^2+5x
=x(x+1)(x+5)
=x(x+1)(x+2+3)
Therefore there must be one out of x, (x+1),(x+5) that is divisible by 3
In other words, x(x+1)(x+5) is a multiple of 3.
On the other hand,
y^3y=y(y1)(y+1)
which is the product of 3 consecutive numbers.
Therefore there must be one that is divisible by 3
Dividing the equation (1) by 3 results in
\dfrac{x(x+1)(x+2+3)}{3} = \dfrac{y(y1)(y+1)}{3} \dfrac{2}{3}
which shows the remainder \dfrac{2}{3} is not an integer. Hence, there's no integer solution to hold the equation equal.