This site is supported by donations to The OEIS Foundation.

# A243100

## Why ${\displaystyle x^{^{y}}-x}$ for ${\displaystyle x,y>1}$ is even?

1) Analyze such difference with the aid of the Newton's binomial in each case for combinations of parity: a) x is even and any y, b) x is odd and y is even, and c) Both odd.

2) To observe that such fact necessarily is true as direct consequence from the definition of "factorial". As follows:

${\displaystyle x^{y}=sum_{k=1}^{min\left(x,y\right)}k!{\begin{Bmatrix}y\\k\end{Bmatrix}}{\binom {x}{k}}}$

By separating the first term (${\displaystyle k=1}$) term:

${\displaystyle x^{y}-x=sum_{k=2}^{min\left(x,y\right)}k!{\begin{Bmatrix}y\\k\end{Bmatrix}}{\binom {x}{k}}}$

It is straightforward to note due the presence of ${\displaystyle k!}$ that the right-hand side is zero mod 2, always that x,y>1.

## Notes

[[For reference: Wikipedia contributors. Stirling numbers of the second kind. Wikipedia, The Free Encyclopedia.][1]]

[[Some related PARI-GP sourcecode about the effect of this result over the definition for A243100][2]]