The integral solution of $x^{2}-y^{3}=1 (x>1,y>1) $? I know it's a special case of catalan's conjecture.Wiki says its only solution is $(3,2)$.But I cannot work  it out.Any help will be appreciated.
 A: You can re-write the expression as $y^3=x^2-1=(x+1)(x-1)$. Now we can do some analysis. The only common factor that $x+1$ and $x-1$ can have is $2$. 


*

*Case 1- They have no common factor. In that case, both $x+1$ and $x-1$ are cubes. This is impossible, as the smallest difference between cubes is $2^3-1^3=7$.

*Case 2- They have one common factor- namely $2$. Then for $(x-1)(x+1)$ to be a cube, $x-1=2a^3$ and $x+1=2^2b^3$, or $x-1=2^2a^3$ and $x+1=2b^3$. Here $a,b$ are positive numbers different from $2$. Hence, $|2a^3-2^2b^3|=2$ or $|2^2a^3-2b^3|=2$. We can now see that the only way that this is possible is that $a=b=1$.


Hence, $x=3, y=2$ is the only solution
A: Are you looking for some specific solution? Do you mean $3^2-2^3=9-8=1$?
A: Not finished!
Write $$(x-1)(x+1)=y^3$$
Case 1. $x$ is even then $x+1$ and $x-1$ are relatively prime so $$ x-1 = a^3$$ $$x+1 = b^3$$ where $a<b$ are relatively prime and $ab=y$. Now we have $$2=(b-a)(b^2+ab+a^2)$$
and so: 


*

*$b-a=1$ and $b^2+ab+a^2 =2$ so $$ 3a^2+3a+1 =2$$ and this is impossibile. 

*$b-a=2$ and $b^2+ab+a^2 =1$ so $$ 3a^2+6a+4=1$$ so $a=-1$ and $b=1$ so $y=-2$ and $x=0$ whivh is not ok again.
Case 2. $x=2z+1$ then $y=2t$ so $$z(z+1)=2t^3$$
Since $z,z+1$ are relatively prime we have


*

*$z=2a^3$ and $z+1=b^3$, so $2a^3=b^3-1= (b-1)(b^2+b+1)$. So $$b-1 = 2p^3$$ $$b^2+b+1=q^3$$ where $p,q$ are relatively prime...

*$z=a^3$ and $z+1=2b^3$, so $2b^3= a^3+1=(a+1)(a^2-a+1)$. So $$a+1 = 2p^3$$ $$a^2-a+1=q^3$$ where $p,q$ are relatively prime...
