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 $\displaystyle a=\frac{y-1+y-2+y-3+y-4}4$
