Find all positive integer solutions of $$ (2x-1)^3 +16 = y^4. $$
I think that there exist no integral solution for this equation, but I am unable to prove it.
Am I right? Is there any solution to this equation, if no how can I prove it?
My attempts: clearly y is odd, so taking congruence: RHS is congruent 1 (mod 4), and LHS is congruent 1 or 3 (mod 4). But this couldn't prove that there's no solution, so I tried taking modulo 3,8,... but got stuck.
Since $(2x-1)$ is odd, it follows that $(2x-1)^3$ is odd too. This forces $y^4$ to be odd, i.e. $y$ is odd too.
Rewrite the given equation as:
\begin{align} (2x-1)^3 & =y^4-16 \\ & = (y^2+4)(y+2)(y-2) \end{align}
Now, we have the following: \begin{align} \gcd(y+2,y-2) & = \gcd(4,y-2) \\ & = 1 \ \text{(Since $y-2$ is odd)} \end{align}
\begin{align} \gcd(y^2+4,y+2) & = \gcd(y^2-y+2,y+2) \\ & = \gcd(y^2-2y,y+2) \\ &= \gcd(y,y+2) \\ &= \gcd(y,2) \\ &=1 \end{align}
To see that $\gcd(y^2+4,y-2)=1$, note that $\gcd(y^2+4,y^2-4)=\gcd(8,y^2-4)=1.$ But since $y^2-4=(y+2)(y-2)$ and $\gcd(y^2+4,y+2)=1$, we are done.
Hence, all $3$ terms on the R.H.S. of our re-written equation are pairwise coprime. If any term is not a perfect cube, then the product of the three terms is also not a perfect cube (can you see why?), contradicting the fact that the L.H.S. of the equation is a perfect cube. Hence, we must have that all 3 terms are perfect cubes. But this is again impossible, since $y+2$ and $y-2$ cannot both be perfect cubes at the same time (again, can you see why?)
We conclude that there are no positive integer solutions to your equation.