How do I find all the integer solutions for $$x^3-y^3-2x^2=1$$ My attempt: I realized that the oddity between $x$ and $y$ need to be different. So I proposed that $x$ is odd and $y$ is even and tried substituting $x=2m+1$ and $y=2n$ and got this equation: $$4m^3+2m^2+m-4n^3=1$$ However, I wans’t able to get more progress. I also tried factoring and got this equation. This gave me $$(y+1)(y^2-y+1)=x^2(x-2)$$ So I found two integer pairs, $(0, -1), (2, -1)$. However, I cannot gurantee this is it. Please help me. Thank you!
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2 Answers
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We have $$(x-1)^3< x^3-2x^2-1 <x^3$$ for all $x\notin \{0,1,2,3\}$
So the only solution for $x$ are in $\{0,1,2,3\}$
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Consider $$(x-1)^3\leq y^3<x^3.$$ The right inequality is true for all integer $x$.
The left inequality is true for $x(x-3)\geq0.$
answered Jul 18, 2019 at 15:13