Is there any integer solutions of $3x^3+3x+7=y^3$?

$3x^3+3x+7=y^3$

$x, y \in \mathbb{N}$

Having thought about it two hours, and I'm still not sure how to show there actually aren't any integer solutions.

EDIT Another formulation of this problem: Prove that $3x^3+3x+7$ cannot be a perfect cube.

Solution 1:

There are no natural solutions for $x$ and $y$.

A quick bit of important information: If $C_n$ is the $n$th non-negative cube ($0,1,8,...$), then $C_{n+1}-C_n = 3n^2+3n+1$. (This can be grasped visually by a similar method to the visual proof that the consecutive differences of consecutive squares are consecutive odd integers.)

Let $s,a$ be positive integers such that $a=s-1$ (this will make the math a bit easier later).

$y$ is clearly greater than $x$ so let $y=x+s=x+a+1$.

Subtract $x^3$ from both sides of $3x^3+3x+7=y^3$ to get $$2x^3+3x+7=y^3-x^3$$

Let us focus on expanding $y^3-x^3$. Note that this is equivalent to $$((x+a+1)^3-(x+a)^3)+((x+a)^3-(x+a-1)^3)+\cdots+((x+2)^3-(x+1)^3)+((x+1)^3-x^3)$$ Which, by expanding the differences of consecutive cubes, is equivalent to: $$3x^2+3x+1+3(x+1)^2+3(x+1)+1+3(x+2)^2+3(x+2)+1+⋯+3(x+a)^2+3(x+a)+1=$$ $$3(x^2+x+(x+1)^2+x+1+\cdots+(x+a)^2+x+a)+a+1=$$ $$3(x^2+(x+1)^2+⋯+(x+a)^2)+3(x+x+1+⋯+x+a)+a+1$$

Note $a+1$ arises since $1$ appears $a+1$ times in the expression.

We now condense the first two terms on the RHS (starting with the first): $$3(x^2+(x+1)^2+⋯+(x+a)^2)=$$ $$3(x^2+x^2+2x+1+x^2+4x+4+x^2+6x+9+⋯x^2+2ax+a^2)=$$ $$3(x^2(a+1)+xa(a+1)+\frac{a(a+1)(2a+1)}{6})$$

These coefficients are obtained by noting there are $a+1$ terms of $x^2$ (as above with $1$s), that the coefficients on the $x$ terms are the first $a$ even integers, and that the constants are the first $a$ squares.

Now we move to the second term on the RHS:

$$3(x+x+1+⋯+x+a)=$$ $$3(x(a+1)+\frac{a(a+1)}{2})$$

Again, the coefficients are found by noting the $a+1$ $x$ terms and the first $a$ positive integers.

Now, returning to $$3(x^2+(x+1)^2+⋯+(x+a)^2)+3(x+x+1+⋯+x+a)+a+1$$ we replace the first two terms with results found above to obtain: $$3(x^2(a+1)+xa(a+1)+\frac{a(a+1)(2a+1)}{6})+3(x(a+1)+\frac{a(a+1)}{2})+a+1$$

At this point it is convenient to substitute $a$ for $s-1$ which leaves us with: $$3(x^2s+xs(s-1)+\frac{s(s-1)(2s-1)}{6})+3(xs+\frac{s(s-1)}{2})+s$$

Distribute the $3$'s: $$3x^2s+3xs(s-1)+\frac{s(s-1)(2s-1)}{2}+3xs+\frac{3s(s-1)}{2}+s=$$ $$3x^2s+3xs^2+\frac{s(s-1)(2s-1)+3s(s-1)}{2}+s=$$ $$3x^2s+3xs^2+\frac{s(s-1)(2s+2)}{2}+3s=$$ $$3x^2s+3xs^2+s(s-1)(s+1)+s=$$ $$3x^2s+3xs^2+s^3$$

UPDATE This answer previously contained a mathematical error which has since been corrected, however as a result it no longer solves the problem.