Let x, y be positive integers such that x^4=(x-1)(y^3-23)-1 . Find...

$$\begin{aligned} & \sqrt{4 n+r}=k \in I \\ & \Rightarrow 4 n+r=k^2 \\ & \frac{\left(x^4+1\right)}{(x-1)}=y^3-23, x, y \in I \end{aligned}$$ $\text { Since, } x \neq 1 \Rightarrow \quad$$\begin{aligned} & x \geq 2 \\ & y^3-23 \in I \end{aligned}$$\Rightarrow \frac{x^4+1}{x-1} \in I$ Let, $x-1=p$\begin{aligned} & \Rightarrow \frac{(1+p)^4+1}{p} \\ & \Rightarrow \frac{\left(a_4 p^4+a_3 p^3+\ldots . . a_1 p+2\right)}{p} \end{aligned}\begin{array}{ll} \Rightarrow & \frac{2}{p} \Rightarrow p \text { divides } 2 \\ \Rightarrow & p=\{-2,-1,1,2\} \\ \Rightarrow & x \in\{-1,0,2,3\} \end{array}but $x \geq 2 \Rightarrow x=2$ or 3\begin{aligned} & \text { If } x=2 \quad 2^4=1 \times\left(y^3-23\right)-1 \\ & \Rightarrow \quad(17+23)=y^3 \\ & \Rightarrow \quad y \notin I \end{aligned}$$If$x=3 \quad\left(3^4+1\right)=2\left(y^3-23\right)=y=4 \Rightarrow \quad x+y=7