Please some hint on how to solve in the set of natural numbers $$x^{100} − y^{100} = 100!$$ The question comes from the Serbian Junior Mathematical Olympiad 2020.
I have tried with Fermat`s theorem using $x^{100}=x^{101-1}$, same with $y$ and that $x$ is a multiple of $101$ and $y^{100}=1 \pmod {101}$
On
I have a clear solution. First, let's apply Fermat's little theorem and we have $ x ^ { 100 } , y ^ { 100 } \equiv 1 \pmod { 101 } $. Then we can use Wilson's theorem and get $ 100 ! \equiv - 1 \pmod { 101 } $. So, the left-hand side of the equation ($ x ^ { 100 } - y ^ { 100 } $) is a multiple of $ 101 $, but the right-hand side ($ 100 ! $) is not. Therefore, no solutions exist in the set of natural numbers.
You have shown that a solution requires $x=101k$. Therefore $$x^{100}-y^{100}\geq101^{100}-100^{100}\geq100\times100^{99}=100^{100}>100!$$
The first inequality follows from $(101k)^{100}-y^{100}$ taking its smallest positive value for fixed $k\in \mathbb{N}$ when $y=101k-1$ and $(101k)^{100}-(101k-1)^{100}$ being an increasing function in $k\in\mathbb{N}$ (and negative when $k=0$).
The second inequality follows from the binomial expansion of $(100+1)^{100}$.
The final inequality follows from $100>1,2,3,\cdots,99$.