Let $$x, y$$ be positive integers such that $$x^{4}=(x-1)(y^{3}-23)-1$$. Find the maximum possible value of $$x+y$$.

The given relation is

$$x^{4}=(x-1)(y^{3}-23)-1 \qquad -(1)$$

Rewrite $$-(1)$$ so that the two factors involving $$x-1$$ become evident:

$$x^{4}+1=(x-1)(y^{3}-23)\qquad -(2)$$

To understand how $$x-1$$ can divide $$x^{4}+1$$, first calculate the greatest common divisor of the two numbers.

Since $$x\equiv 1 \pmod{x-1}$$, we have $$x^{4}\equiv 1 \pmod{x-1}$$, hence

$$x^{4}+1\equiv 2 \pmod{x-1}\,.$$

Therefore

$$\gcd(x-1,\;x^{4}+1)\in\{1,2\}\,.$$

Next perform polynomial division of $$x^{4}+1$$ by $$x-1$$:

$$x^{4}+1=(x-1)(x^{3}+x^{2}+x+1)+2 \qquad -(3)$$

Combine $$-(2)$$ and $$-(3)$$:

$$(x-1)(y^{3}-23)=(x-1)(x^{3}+x^{2}+x+1)+2$$

which simplifies to

$$(x-1)\bigl[y^{3}-23-(x^{3}+x^{2}+x+1)\bigr]=2 \qquad -(4)$$

Equation $$-(4)$$ shows that $$x-1$$ divides $$2$$, so

$$x-1\in\{1,2\}\quad\Longrightarrow\quad x\in\{2,3\}\,.$$

Substitute in $$-(1)$$:

$$2^{4}=(2-1)(y^{3}-23)-1$$ $$16 = 1\cdot(y^{3}-23)-1$$ $$y^{3}=40$$

Since $$40$$ is not a perfect cube, no integer $$y$$ exists in this case.

Substitute in $$-(1)$$:

$$3^{4}=(3-1)(y^{3}-23)-1$$ $$81 = 2(y^{3}-23)-1$$ $$2(y^{3}-23)=82$$ $$y^{3}=64$$ $$y=4$$

This is an admissible positive-integer solution.

The only solution in positive integers is $$(x,y)=(3,4)$$, giving

$$x+y=3+4=7\,.$$

Hence the maximum possible value of $$x+y$$ is 07.