Ok here i present my proof: please point out any flaws .. $(y-1)(y^2+y+1)=(x^2)(x^2+1)$ $x^2$ and $x^2+1$ are coprime now u only need to do some nasty casework

That leads nowhere. Having $A\cdot B = C\cdot D$, with $C,D$ coprime, even for $A,B$ coprime is useless (you cannot simply equate $A=C, B=D$ or vice-versa).

digantabhattacharya2 wrote: Ok here i present my proof: please point out any flaws .. $(y-1)(y^2+y+1)=(x^2)(x^2+1)$ $x^2$ and $x^2+1$ are coprime now u only need to do some nasty casework Though I have no idea how you would proceed after this, there are no flaws in the factorisation that you did. I am pretty sure of that. WLOG, $x$ is positive. And obviously, $y$ is positive too. \[ y^3-1=x^4+x^2 \implies y^3=x^4+x^2+1 =(x^2+x+1)(x^2-x+1) \] Notice now that $(x^2+x+1,x^2-x+1)=(x^2+x+1,2x)$. Since $x^2+x+1$ is odd, we have this quantity $=(x^2+x+1,x)=1$. Therefore, both of $x^2+x+1$ and $x^2-x+1$ are cubes. Let $x^2+x+1=a^3$ and $x^2-x+1=b^3$. Take $2x+1=m$ and $2x-1=n$. Then, $m^2+3=4a^3$ and $n^2+3=4b^3$ and $m=n+2$. Obviously, $a \ge b+1$. Therefore, \[ 4a^3-4b^3=(n+2)^2-n^2=4n+4 \implies a^3-b^3=n+1 \] But, \[ a^3-b^3 \ge (b+1)^3-b^3=3b^2+3b+1 \ge 3b^2 = 3 \left( \frac{n^2+3}{4} \right)^{2/3} \]So, we have $n+1 \ge \left( \frac{n^2+3}{4} \right)^{2/3} \implies 16(n+1)^3 \ge (n^2+3)^2$. Its easy to show using induction that the only positive integer solutions to this inequality are $0 \le n \le 18$. From here, we get $x \le 9$ Now its just case work. There are no solutions. EDIT: I forgot to mention the trivial solution $(x,y)=(0,1)$.

mathisfun7 wrote: Find all pairs $(x,y)$ of integers such that $y^3-1=x^4+x^2$. $y^3 = (x^2+x+1)(x^2-x+1)$. Since $(x^2+x+1,x^2-x+1)=1$ We must have both as perfect cubes. Thus take $x^2-x+1 = m^3$ and $x^2+x+1 = n^3$. Now $x^2-x+1 = m^3 \le (n-1)^3 < n^3 -3n^2+3n = x^2+x+1 - 3n^2+3n $ . Since $n^3 > x^2$, and if $n \ge 3$, we must have $x^2+x+1 -3n^2+3n \le x^2+x+1-2n^2 \le x^2+x-1 -2x^{4/3}$.. We have no solutions if $n \le 2$. And since $x^2+x-1 -2x^{4/3} \ge x^2-x+1 $ which gives $(x,y)=(0,1)$ So we are done! $\Box$

arkanm wrote: mathisfun7 wrote: This means that $k^2+k=z^2+z$, or $(k-z)(k+z)=-(k-z)$. Hence either $k=z$ or $k+z=0\Rightarrow k=-z$. The first case sends us in circles, so $k=-z$. This is not sending us in circles - yours is a circular reasoning. In effect from $y^3-1 = x^4 + x^2$ you arrived at $y^3-1 = k^2 + k = z^2+z$, where you denoted $z:=x^2$. This is just an identity, for $k=z$. Not to mention that if we assume $k\neq z$, it will follow $k+z=-1$, not $k+z=0\Rightarrow k=-z$, as you wrote, and the rest is suffering. It's a miracle you thus reached the solution $(x,y) = (0,1)$.

Firstly notice that $y$ is positive. Also if $x=0$, then $y=1$. WLOG, $x$ is also positive. Note that the given is actually equivalent to $$y^3=(x^2+x+1)(x^2-x+1).$$Also we have $\gcd(x^2+x+1,x^2-x+1)=\gcd(x^2-x+1,2x)=1$. Thus, $$ \begin{cases} x^2+x+1=a^3\\ x^2-x+1=b^3, \end{cases} $$for some positive integers $a>b$. Hence, $$\left(\frac{a^3-b^3}{2}\right)^2+1=\frac{a^3+b^3}{2}\Longleftrightarrow (a-b)^2(a^2+ab+b^2)^2+4=2a^3+2b^3.$$However as $a\geq b+1$, we get $$LHS\geq (a^2+ab+b^2)^2+4>2a^3+2b^3=RHS$$for any positive $a,b$. No solutions here. We conclude that our only solution is $\boxed{(x,y)=(0,1)}$.