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I began the proof by factoring the LHS to be $x(x+1)+1 = y^2$. Since the LHS contains the product of two consecutive integers, that part is even. But since $1$ is added, it must be odd. That means $y$ must also be odd. How should I continue from here?

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Hint:

If $x^2+x+1=y^2$ with $x$ a positive integer, then $x^2<y^2<(x+1)^2=x^2+2x+1$

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  • $\begingroup$ That makes sense. I don't understand where you are going though, could you please explain? $\endgroup$ – Albert Chung Sep 11 '19 at 23:49
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    $\begingroup$ there are no perfect squares between $x^2$ and $(x+1)^2$, so $y$ can't be an integer; i.e., $y$ is strictly between consecutive integers $x$ and $x+1$, so $y$ is not an integer $\endgroup$ – J. W. Tanner Sep 11 '19 at 23:54
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    $\begingroup$ Proof by contradiction: suppose the equation has a solution in the positive integers. However, using Tanner's analysis, you can see that $x < y < x + 1$, so $y$ cannot be an integer, and there's your contradiction. Done. $\endgroup$ – Sigma Sep 11 '19 at 23:58
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$x^2+x+1=y^2$ implies $4x^2+4x+4=(2y)^2$ or $(2y)^2-(2x+1)^2 = 3$. The only integer squares that differ by $3$ are $4$ and $1$, so the only solutions are given by $(x,y)\in\{(0,1),(0,-1),(-1,1),(-1,-1)\}$.

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