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 2020-02-26, 14:35 #1 enzocreti   Mar 2018 10000011112 Posts Two variable functions 223*34^x+5879*y-(y-2)*17^2+10=f(x,y) 23*34^x+5879*y-(y-2)*17^2+10=g(x,y) I search integers k>0 such that either f(x,y)=k or g(x,y)=k have integer solutions x and y greater than 0. Is it possible to find such k up to 10^6? Last fiddled with by enzocreti on 2020-02-26 at 14:46
2020-02-27, 04:52   #2
CRGreathouse

Aug 2006

10111010010012 Posts

Quote:
 Originally Posted by enzocreti Is it possible to find such k up to 10^6?
It is! I find 678 such members. You could, too, with my code below and a copy of PARI/GP. The parameters are pretty extensible in case you'd like to modify things.

Code:
liste(lim,mul,base=34,step=5590,offset=588)=my(v=List(),X=mul*base); lim\=1; while(X<lim,forstep(n=X+offset+step,lim,step, listput(v,n)); X*=base); Set(v)
list(lim)=setunion(liste(lim,23),liste(lim,223))
v=list(1e6)
Up to a billion I find 1370532 or 0.1%. Asymptotically I think the density is about 1.5% (42/2795) but it will take a while to get there.

2020-02-27, 06:13   #3
enzocreti

Mar 2018

17·31 Posts

Quote:
 Originally Posted by CRGreathouse It is! I find 678 such members. You could, too, with my code below and a copy of PARI/GP. The parameters are pretty extensible in case you'd like to modify things. Code: `liste(lim,mul,base=34,step=5590,offset=588)=my(v=List(),X=mul*base); lim\=1; while(X

Among the solutions there are k=69660, 92020 and 541456 which are the exponents of pg primes multiple of 86.
Is it a chance?

2020-02-27, 07:17   #4
CRGreathouse

Aug 2006

3×1,987 Posts

Quote:
 Originally Posted by enzocreti Among the solutions there are k=69660, 92020 and 541456 which are the exponents of pg primes multiple of 86. Is it a chance?
Probably! You cover a lot of residue classes, and your method isn't clear. Had you specified a method *before* I found the list it would have been more compelling.

2020-02-27, 08:15   #5
enzocreti

Mar 2018

17·31 Posts
...

Quote:
 Originally Posted by CRGreathouse Probably! You cover a lot of residue classes, and your method isn't clear. Had you specified a method *before* I found the list it would have been more compelling.

All the k's (for k multiple of 43) that are solutions if I am not wrong are congruent to (7^3+1) mod 559.
Why?

Last fiddled with by enzocreti on 2020-02-27 at 09:00

2020-02-27, 10:01   #6
enzocreti

Mar 2018

17·31 Posts
...

Quote:
 Originally Posted by CRGreathouse Probably! You cover a lot of residue classes, and your method isn't clear. Had you specified a method *before* I found the list it would have been more compelling.

92020, 69660 and 541456 are moreover congruent to 10^m mod 41...

I don't know if there are other k's solutions that are congruent to 10^m mod 41

2020-02-28, 02:21   #7
CRGreathouse

Aug 2006

174916 Posts

Quote:
 Originally Posted by enzocreti 92020, 69660 and 541456 are moreover congruent to 10^m mod 41... I don't know if there are other k's solutions that are congruent to 10^m mod 41
I mean, znorder(Mod(10,41)) = 5. It's not a rare property.

2020-02-28, 08:50   #8
enzocreti

Mar 2018

10178 Posts
...

Quote:
 Originally Posted by CRGreathouse I mean, znorder(Mod(10,41)) = 5. It's not a rare property.

yes I don't understand why the k solutions multiple of 43 are all congruent to 344 mod 559

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