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This is what I have so far:

$$x^3-x = 12y + 6$$ $$x(x+1)(x-1) = 2(6y+3)$$ The RHS of the equation is even, so therefore so must the LHS. Given that the three numbers on the LHS are consecutive, then we know $x$ is even and $(x-1)$ and $(x+1)$ are odd.
Given that they're three consecutive numbers their gcd is 1, making them coprime.

That's as far as I got. Please note that both $x$ and $y$ are positive integers.

Steven01123581321
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BobMarley
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    Hint: Consider the solutions of the congruence $x^3-x-6\equiv0$ mod $4$. – Barry Cipra Feb 26 '21 at 12:42
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    What's wrong with odd $x$ and even $x\pm1$? – Andrew Chin Feb 26 '21 at 12:44
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    Rewrite equation as $\binom{x+1}{3}=2y+1$. – user10354138 Feb 26 '21 at 12:45
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    @AndrewChin, what's the difference between odd $x$ and even $x\pm1$? – Barry Cipra Feb 26 '21 at 12:54
  • @user10354138, nice observation, but I'm not sure how much it helps find values of $x$ that give integer values for $y$. Do you have something binomial or Pascalian in mind? – Barry Cipra Feb 26 '21 at 12:57
  • @BarryCipra: user10354138's hint shows that an integer solution for $y$ exists when $\binom{x+1}3 = \frac {(x-1)(x)(x+1)}{6}$ is odd, which can be solved mod 4. – player3236 Feb 26 '21 at 13:11
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    @player3236, yes, but what you describe effectively just undoes the introduction of the binomial notation (and even the division by $6$), and gets back to what I suggested doing in my initial comment. So I still don't see how rewriting the equation as suggested shines any helpful light on how to go about solving it for integer values of $y$. – Barry Cipra Feb 26 '21 at 15:17

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Note that necessarily $x\equiv2\pmod{4}$ because $$x(x-1)(x+1)=12y+6\equiv2\pmod{4},$$ and conversely if $x=4z+2$ then $$x^3-x=4(16z^3+24z^2+5z)+6,$$ where a quick check shows that $16z^3+24z^2+5z\equiv0\pmod{3}$ for all $z$.

Servaes
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