Prove or disprove: $99^{100}+100^{101}+101^{99}+1$ is a prime number

Question

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.

Answer 1

Here is a simple Python script to calculate $a^b$ mod $p$:

def powmod(a, b, p):
  if b==0: return 1
  tmp = powmod(a, b/2, p)**2
  if b%2!=0: tmp *= a
  return tmp%p

To check divisibility of your function by $p$, you can use:

def ck(p):
  val = powmod(99,100,p) + powmod(100,101,p) + powmod(101,99,p) + 1
  return (val%p == 0)

Finally, testing all the numbers through $10^6$ for divisibility:

>>> filter(ck, xrange(2,1000000))
[825277]

So it's not prime.

Answer 2

To promote Sage:

sage: a = 99^100 + 100^101 + 101^99 + 1
sage: N = 10^100
sage: for p in primes(N):
....:     if a % p == 0:
....:         print p
....:         break
....:         
825277

Answer 3

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