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.

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: 


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*$b-a=1$ 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...
