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

[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?

 

[fresh_42]

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?

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

 

[mfb]

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

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?

 

[Janosh89]

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

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)...

 

[Janosh89]

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?

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

 

[fresh_42]

It is not so trivial if you want to consider all factors of the number.

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

 

[Janosh89]

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?

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

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 )

 

[Janosh89]

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

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

 

[SlowThinker]

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 )

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.

 

[mfb]

For the original question, the first step is to write the equation modulo 10.

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.

 

[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.

 

[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 )
These have the same factors , being derived from 20x^2-1 !

 

[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

 

[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

 

[mfb]

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.

 

[Janosh89]

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