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#1 |
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Mar 2018
10228 Posts |
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 |
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#2 |
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Aug 2006
10111010110112 Posts |
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) |
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#3 | |
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Mar 2018
2·5·53 Posts |
Quote:
Among the solutions there are k=69660, 92020 and 541456 which are the exponents of pg primes multiple of 86. Is it a chance? |
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#4 |
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Aug 2006
135338 Posts |
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.
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#5 | |
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Mar 2018
10000100102 Posts |
Quote:
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 |
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#6 | |
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Mar 2018
10228 Posts |
Quote:
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 |
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#7 |
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Aug 2006
135338 Posts |
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#8 |
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Mar 2018
2×5×53 Posts |
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