48

How would you go about finding prime factors of a number like $7999973$? I have trivial knowledge about divisor-searching algorithms.

Asaf Karagila
  • 370,314
  • 41
  • 552
  • 949
  • 10
    This is the only context you have to solve the problem. Find a prime factor of 7999973 without a calculator. I got this problem from an old problem book, but there is no explanation or further context. –  Jan 21 '18 at 23:52
  • 4
    Perhaps you didn't read the question carefully... or perhaps you didn't write the question carefully. Finding prime factors (implicitly: all prime factors) of the number, without a calculator, is going to be quite hard. Finding **a** prime factor (that is: only one asked for), as in your title, is quite easy in this case. – David Jan 21 '18 at 23:57
  • @David: `7999973` is the product of two primes. Finding one prime factor will give you the other. You're right though, it will be hard to prove that the largest one is prime. – Eric Duminil Jan 22 '18 at 14:08
  • 4
    If ever I could understand what is special with this question and such votes – Guy Fsone Jan 22 '18 at 21:27
  • 2
    Those who don't understand this problem might understand it more if they assumed that it's a problem that is meant to be solvable in a reasonable amount of time. When you take that approach, it's safe to assume there must be some trick or non-linear thinking, as the answer confirms. – Todd Wilcox Jan 22 '18 at 21:34
  • @EricDuminil That's exactly my point. There is no way **easily and without a calculator** to know that this number is a product of two primes. – David Jan 22 '18 at 22:54
  • 1
    Possible duplicate of [How to factorise large number without calculator?](https://math.stackexchange.com/questions/2309238/how-to-factorise-large-number-without-calculator) – Rob Jan 23 '18 at 10:55
  • As indicated in the other comments, if you are expected to do it by hand in a reasonably short amount of time (e.g. on a single piece of paper) then there must be some special property of the number that makes it easy , In this case it's the difference of 2 cubes. In general you need to test divisibility of $n$ by every prime not exceeding $\sqrt n\;$ to find whether $n$ is, or is not, prime. – DanielWainfleet Jan 24 '18 at 05:35

1 Answers1

210

The thing to notice here is that 7,999,973 is close to 8,000,000. In fact it is $8000000 - 27$. Both of these are perfect cubes. Differences of cubes always factor: $$a^3 - b^3 = (a-b)(a^2+ab+b^2)$$

Here we have $a=200, b=3$, so $a-b= 197$ is a factor.

MJD
  • 62,206
  • 36
  • 276
  • 489
  • 3
    And Pari says the other is $40609$. – Bernard Jan 22 '18 at 00:04
  • 13
    @Bernard: Also note that this method does not prove that the factors we got are prime. – user21820 Jan 22 '18 at 09:30
  • 13
    @user21820 As clarified in a comment, the original question only required finding *a* prime factor, and 197 is small enough to check by hand. – Geoffrey Brent Jan 22 '18 at 10:26
  • 6
    @GeoffreyBrent: Yup I know that 197 is prime; I'm just stating for the record that this method in general does not say anything about primality. – user21820 Jan 22 '18 at 10:33
  • 2
    @user21820 well it gives you a relatively small factor (compared to 7.999.973), and then you could still divide this one by any of its factors until you got a prime – Rafalon Jan 22 '18 at 15:14
  • 9
    @Rafalon trying to divide by 7, 11 and 13 is enough (for 197). – Will Ness Jan 22 '18 at 20:02
  • 1
    @WillNess that's exactly my point :) – Rafalon Jan 22 '18 at 21:44
  • 1
    @MJD Love your answer but why exactly did you choose 8 million as your starting point.. Too satisfy the two wo perfect cubes ? or a deeper theory? – Peter H Jan 23 '18 at 06:09
  • 4
    @PeterH Because the original number was suspiciously close to $8\,000\,000$, with all those $9$'s, so he checked to see whether it worked, and it did. It's a standard type of problem, although in my experience, powers of $10$ are more common as starting points for these problems than numbers like eight million. – Arthur Jan 23 '18 at 09:54