# B Why is the quadratic expression 20*x^2-1 only divisible by 11,19,29...

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1. Apr 13, 2017

### Janosh89

Ian in London (South).Interests Number Theory. Prime Numbers...
Integer iterations of this quadratic expression only yield decimal numbers whose factors end with the digit _1, or _9. Why?

2. Apr 13, 2017

### Staff: Mentor

Do you have any idea? Without the "-1", which is the last digit?

3. Apr 13, 2017

### Staff: Mentor

It is not so trivial if you want to consider all factors of the number. 13*23 can give the correct last digit, for example.

@Janosh89: Can you rule out some factors based on modular arithmetic?
Can you generalize that?

4. Apr 13, 2017

### Janosh89

The factors are in the set 11,19,29,31,41,59,61,71 and so on but can include prime numbers (^2), (^4),(^6)...

Last edited by a moderator: Apr 14, 2017
5. Apr 13, 2017

### Janosh89

299 (13*23) is not expressible by 20*x^2-1 ? where x is an integer

6. Apr 13, 2017

### Staff: Mentor

Yes, I erroneously thought it was about the digits of $20x^2-1$ itself.
Interesting question though and indeed not trivial.

7. Apr 14, 2017

### Janosh89

This is applicable to a sub-class of prime pairs, 59;61 say ,
45x^2 +15x +1
45x^2 +15x - 1 ( 5x^2+5x +/-1 where x=3*n )

8. Apr 14, 2017

### Janosh89

Hope you got my reply. Thanks very much for responce to my initial question.

9. Apr 14, 2017

### SlowThinker

I don't understand what this means. Is it another question, or an attempt at answering the original question?
For the original question, the first step is to write the equation modulo 10.

10. Apr 14, 2017

### Staff: Mentor

And how does that help?

The obvious approaches don't directly lead to a solution. While it is easy to rule out many specific primes as factors, that is not trivial to generalize.
This is an interesting problem. Please be careful with hints if you don't know if they help.

11. Apr 14, 2017

### SlowThinker

I think I have a proof, I just didn't want to post it.
In the first step I prove that factors must be 1,3 or 9 mod 10. In the second step I rule out those with 3 mod 10.

Edit: I guess I'll have to rule out 7 mod 10 as well...
Edit2: Yes the first step allows 7 mod 10 as well but second step can be repeated to rule those out as well.

Last edited: Apr 14, 2017
12. Apr 15, 2017

### Janosh89

13. Apr 15, 2017

### Janosh89

Thanks for signposting! Maybe it is more trivial, but
60*x^2+1 has only factors in the sets

30*n+1, +17, +19, +23

14. Apr 18, 2017

### Janosh89

45x2+15x +/-1. ... 59;61 ,209;211 ,449;451 ,779;781 ...
45x2-15x +/-1 ... 29;31 ,149;151 ,359;361 ,659;661 ....
have the same factors as 20x2-1

15. Apr 18, 2017

### Staff: Mentor

45*1383768138376181^2-15*1383768138376181-1 = 61 × 1 412567 897300 469698 405475 260189
45*1383768138376181^2-15*1383768138376181+1 = 11 × 7 833331 066848 059236 612180 988321
45*1383768138376181^2+15*1383768138376181-1 = 547 101119 × 157496 738249 823782 787361
45*1383768138376181^2+15*1383768138376181+1 = 86 166641 735328 693115 778142 156961 (prime)

45*332222111111111111113837681383761^2-15*332222111111111111113837681383761-1 = 41 × 15031 × 473798 391866 453915 014639 × 17 009998 181762 817836 249229 315400 436941
45*332222111111111111113837681383761^2-15*332222111111111111113837681383761+1 = 79 × 101 582219 × 2 051039 191429 × 301 752471 096290 191134 883798 553758 009803 236039
45*332222111111111111113837681383761^2+15*332222111111111111113837681383761-1 = 11 × 451519 900000 050505 057916 361767 462680 924222 331768 767051 457652 135169
45*332222111111111111113837681383761^2+15*332222111111111111113837681383761+1 = 4 966718 900000 555555 637079 979442 089490 166445 649456 437566 034173 486861 (prime, 67 digits)

It certainly generates large primes!

Edit: Found an alternative way to factorize the one nasty case.

Last edited: Apr 18, 2017
16. Apr 18, 2017

### Janosh89

You've made my evening. See you soon. Great!