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I am trying to solve solve $x^2 + x = y^4 + y^3 + y^2 + y$ over the integers. So far I have decomposed it into $x(x + 1) = y(y + 1)(y^2 + 1)$ and noticed that both sides of the equation are nonnegative. Furthermore $GCD(x, x + 1) = GCD(y, y + 1) = GCD(y, y^2 + 1) = GCD(y + 1, y^2 + 1) = 1$.

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Hint: If you multiply both sides by $4$ and add $1$, you get

$$(2x+1)^2=4y^4+4y^3+4y^2+4y+1=(2y^2+y+1)^2-(y^2-2y).$$

This results in two squares that are very close together. Can you prove that they are too close together when $y$ is sufficiently large?

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  • $\begingroup$ Sorry, I don't understand. $\endgroup$ – user128409235 Feb 1 '18 at 17:47
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    $\begingroup$ LHS is a square, and so should be RHS. $(2y^2+y+1)^2$ is a square, but your RHS is not quite that; it is smaller. Is it small enough to be another square? $\endgroup$ – Ivan Neretin Feb 1 '18 at 18:19
  • $\begingroup$ So there is no need to check $y$ values such that $(2y^2 + y + 1)^2 - (2y^2 + y)^2 > y^2 - 2y > 0$. $\endgroup$ – user128409235 Feb 1 '18 at 18:46
  • $\begingroup$ @user128409235 Yes. Most $y$ values will have this property - once you prove which you can just do a finite case check. $\endgroup$ – Carl Schildkraut Feb 1 '18 at 19:37

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