# How to solve $(x-2)^4+(x-3)^4=1$?

I'm trying my hand on these types of expressions. how to solve $x$ and $y$ in $\displaystyle(x-2)^4+(x-3)^4=1$ and $\displaystyle (y-1)(y-2)(y-3)(y-4)=1$?
please write step by step solution ! thank u

HINT: For the first case set $$z=\frac{x-2+x-3}2\iff 2z=2x-5$$

$$2^4=(2x-4)^2+(2x-6)^2=(z+1)^4+(z-1)^4=2\left(z^4+1+\binom42z^2\right)$$

Rearrange to form a Quadratic Equation in $z^2$

For the second set $\displaystyle a=\frac{y-1+y-2+y-3+y-4}4$

• what is z and a here? and please explain more! – jackcall Apr 3 '14 at 15:02
• @user139567, See the definition of $z$ at the first line – lab bhattacharjee Apr 3 '14 at 15:04
• ok but i dont understand why did u write that ??! – jackcall Apr 3 '14 at 15:08
• @user139567, obsrev that in either cases, I've replaced with the mean of the terms; it will eliminate the terms containing odd degree of $z$ and $a$ respectively – lab bhattacharjee Apr 3 '14 at 15:10
• sorry but im not good at english .can u explain clearly please ! – jackcall Apr 3 '14 at 15:29

HINTS

case 1: $2$ and $3$ are obvious solutions. Divide the polynom by $(x-2)(x-3)$ and solve the result.

case 2: $(y-1)(y-2)(y-3)(y-4) = [(y^2−5y)+5]^2−1$ see this post about (y-1)(y-2)(y-3)(y-4)