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

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.

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

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