<s>[INST] 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.
 [/INST] 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
</s>