I came across this problem:
Let x, y be positive integers such that $ x^4 = (x-1)(y^3-23) - 1 $. Find maximum possible value of $ x + y $
which I tried to solve using Lagrange Multipliers:
$ f(x,y) = x + y $
$ g(x, y) = x^4 - (x-1)(y^3-23) + 1 $
$ \frac{\partial f(x,y)}{\partial x} + \lambda\frac{\partial g(x,y)}{\partial x} = 0$
$ \implies 1 + \lambda[ 4x^3 - (y^3-23)] = 0 \quad\to \ (i)$
$ \frac{\partial f(x,y)}{\partial y} + \lambda\frac{\partial g(x,y)}{\partial y} = 0$
$ \implies 1 + \lambda[ -3y^2(x-1) ] = 0 \quad\to \ (ii)$
$ g(x,y) = 0$
$ \implies x^4 - (x-1)(y^3-23) + 1 = 0 \quad\to \ (iii)$
However with this, I couldn't reach to a solution, plotting the curves didn't benefit either. What crucial step am I overlooking?
The given answer is: $x+y=7$
Remark: First, this problem cannot be solved with Lagrange Multiplier technique as $(x,y)\in \mathbb{N}_+^2$.
We have: $$(x-1)(y^3-23)= x^4 +1 =(x^4-2x^2+1)+2x^2=(x^2-1)^2 + 2x^2 = (x-1)^2(x+1)^2 +2x^2$$ We deduce then $(x-1) |2x^2$. As the greatest common divisor of $(x-1)$ and $x$ is $1$, we must have $$x-1 |2$$ $$\implies \cases{x-1= 1\\x-1=2} \iff\cases{x= 2\\x=3}$$ If $x = 2$ then $2^4 = y^3 - 23 -1 \implies y \not \in \mathbb{N}$.
If $x = 3$ then we have : $3^4 = (3-1)(y^3 -23) -1 \iff y = 4$.
So, with the constraint $x^4 = (x-1)(y^3-23)-1 $, the maximum value (or even the only value) of $x + y$ is $ 7$.