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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.

VividD
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math110
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    It had better not be a prime number, because otherwise that would make this a very very difficult problem to prove... – Dustan Levenstein Mar 17 '14 at 15:04
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    Note that $$(x -1)^{x-1} = (100^{101} - 1)^{100^{101}-1}$$ similarly $(x+1)^{x-2} = 101^{99}$ is wrong – dani_s Mar 17 '14 at 15:06
  • Wolfram says it isn't although you will need a pro account to check the prime decomposition. Also this is an approximately 200 digit number. Too large for a normal computer to factorise but I would say that some sufficiently elaborated algorithms with a powerful enough computer should be able to factor it. 3 years ago a classmate factored an 180 digits number in an RSA encryption exercise. – user88595 Mar 17 '14 at 15:09
  • Here's an exercise for you: write up a quick script to compute this number modulo an input prime. ;-) – Dustan Levenstein Mar 17 '14 at 15:11
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    Divisible by 825277. – Chen Wang Mar 17 '14 at 15:12
  • I guess it is divisible by $11$? – dani_s Mar 17 '14 at 15:12
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    @Chen Wang how did you compute that? – dani_s Mar 17 '14 at 15:14
  • @dani_s I think it's 8 mod 11. – Dustan Levenstein Mar 17 '14 at 15:14
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    @dani_s Alpertron's applet. – Chen Wang Mar 17 '14 at 15:15
  • @china math where did you get this problem from? – dani_s Mar 17 '14 at 15:33
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    OP should state that this is a "no calculators allowed" problem, if it even is one. – Yiyuan Lee Mar 17 '14 at 15:35
  • @YiyuanLee well of course, no one here is going to rest till we get a good answer. helps to know, what we have to prove. I am thinking along the lines of factor theorem maybe. – Guy Mar 17 '14 at 15:40
  • @dani_s,this problem is my frend ask me,maybe is Maths Olympic question – math110 Mar 17 '14 at 15:40
  • @chinamath okay. makes sense for it to be in an olympiad. – Guy Mar 17 '14 at 15:42
  • @chinamath could you edit the post to say something like "we have found a factor 825277. so it is not prime, but i am looking for proof without calculator"? – Guy Mar 17 '14 at 15:45
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    I've looked at $$k\mapsto k^{k+1}+(k+1)^{k+2}+(k+2)^k+1$$ with the aim of proving the claim by induction. Didn't work, but here some thoughts: The map preserves the property of being even/odd. The result is prime for $k=1$ and $k=3$, but isn't prime for the next 1000 $k$'s I've checked computationally. I couldn't see any pattern in the prime factors though, they are big and jump in value. Code in Mathematica is 'FactorInteger /@ Table[k^{k + 1} + (k + 1)^{k + 2} + (k + 2)^k + 1, {k, 1, 33}]' – Nikolaj-K Mar 17 '14 at 16:05
  • @NikolajK. nice property about being preserving even-odd. was obvious, yet didn't notice. – Guy Mar 17 '14 at 16:08
  • our number is $k=99$? yes? – Guy Mar 17 '14 at 16:08
  • @NikolajK. don't have mathematica. can you post a link to a text file or something? – Guy Mar 17 '14 at 16:09
  • @Sabyasachi: A text file of what? – Nikolaj-K Mar 17 '14 at 16:12
  • @NikolajK. list of prime factors. of the 1000k's you have checked. but on second thought. nevermind. not much use i figure. – Guy Mar 17 '14 at 16:12

3 Answers3

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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.

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

robjohn
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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
Christoph
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  • Shouldn't N = 11 * 10^100 or something similar as it needs to be at least $\sqrt a$ and $a$ is a little bigger than $100^{101}$ – Henry Jun 21 '25 at 00:19