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Edit: Thank you for the help. It makes much more sense now.

I'm trying to learn proofs on my own, and I would really appreciate any feedback for my proof below. Are there any logical errors? Is there a better way to do this? thanks very much in advance.

Prove that there are no positive integer solutions for $x^2+x+1=y^2$.

My proof: Suppose there exists a pair of integers $x$ and $y$ such that $x^2+x+1=y^2$.

If $y\in Z$, then $y^2$ is a perfect square. It then follows that $x^2+x+1$ is also a perfect square, so can be written in the form $(x+a)^2$ for some $a \in Z$. Then $$x^2+x+1=(x+a)^2=x^2+2ax+a^2.$$ Comparing coefficients gives both $a=1$ and $a=\frac{1}{2} \notin Z$. Therefore $x^2+x+1$ is not a perfect square because it cannot be written in this form. Thus $y^2$ is not a perfect square.

Hence $y$ is not an integer and this contradicts the assumption that $x$ and $y$ are integers for $x^2+x+1=y^2$.

Mari
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    The "comparing coefficients" step is not justified. Two polynomials in $x$ have the same coefficients if they describe the same function of $x$, but not necessarily if they just happen to be equal for one value of $x$. – Karl Aug 02 '22 at 17:33
  • You're on the right track, though. What's the next largest square number after $x^2$? – Karl Aug 02 '22 at 17:42
  • Is it $(x+1)^2$? And this isn't equal to $x^2+x+1$, and if a > 1 then the coefficient of x just gets further away from 1, and there is no way to write the polynomial as a perfect square? – Mari Aug 02 '22 at 17:55

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$$x^2<x^2+x+1 =y^2<(x+1)^2$$

Since $x^2$ and $(x+1)^2$ are two consecutive perfect squares, $x^2+x+1$ is not a perfect square.

There is no positive integer solution.

But for $x=0$, $y=\mp 1$

Lion Heart
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