TITLE: Prove or disprove: $99^{100}+100^{101}+101^{99}+1$ is a prime number
QUESTION [13 upvotes]: Prove or disprove: $$99^{100}+100^{101}+101^{99}+1$$ is a prime number.

My idea: let $100^{101}=x^{x+1}$,then
$$99^{100}+100^{101}+101^{99}+1=(x-1)^{x}+x^{x+1}+(x+1)^{x-1}+1$$ is prime number？
I have solved following problem before:

Show that $$5^{100}+5^{75}+5^{50}+5^{25}+1$$ is not prime number.

Let $x=5^{25}$, then
$$x^4+x^3+x^2+x+1=(x^2+3x+1)^2-5x(x+1)^2$$
Note that
$$5x(x+1)^2=5^{26}(x+1)^2=[5^{13}(x+1)]^2$$
But for my problem it looks I can't use this approach.

REPLY [3 votes]: Similar Mathematica code is FactorInteger[99^100 + 100^101 + 101^99 + 1, 2] which returns $825277$ and $\frac{99^{100} + 100^{101} + 101^{99} + 1}{825277}$.