How many integer solution sets $$(x,y)$$ for the equation $$x^3+6x^2+5x=y^3-y+2$$?

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Uniteasy Admin · Accepted Answer

$$x^3+6x^2+5x=y^3-y+2 $$n
$$x^3+6x^2+5x$$
$$=x(x+1)(x+5)$$
$$=x(x+1)(x+2+3)$$
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,
$$y^3-y=y(y-1)(y+1)$$
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
$$\dfrac{x(x+1)(x+2+3)}{3}  = \dfrac{y(y-1)(y+1)}{3} -\dfrac{2}{3}   $$
which shows the remainder $$-\dfrac{2}{3}$$ is not an integer. Hence, there's no integer solution to hold the equation equal.

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