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Old 2020-02-10, 16:03   #1
enzocreti
 
Mar 2018

17×31 Posts
Default 51+163k

numbers of the form 51+163k






conjecture


51+163k is a square only for k=10
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Old 2020-02-10, 17:04   #2
Batalov
 
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"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

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Code:
for(k=1,10^10,if(issquare(51+163*k),print(k)))
10
91
255
498
826
1231
1723
2290
2946
3675
4495
5386
6370
7423
8571
9786
11098
12475
13951
15490
17130
18831
20635
22498
24466
26491
28623
30810
33106
...
...
Yes infinitely many lol

Last fiddled with by enzocreti on 2020-02-10 at 17:34
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Old 2020-02-10, 23:39   #3
NHoodMath
 
Jan 2017

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Enzo, here is some relevant maths to what you have posted here:

The number 51, in your case, you were testing to see whether it was a quadratic residue (mod 163), that is, whether or not there is a square that is 51 (mod 163). There is a well-known theorem that states that if a number b is a quadratic residue (mod N), then there are infinitely many squares that are b (mod N). In this case, you found that for k=10, there is a square of the form k*N+b for N=163, b=51, therefore 51 is a quadratic residue (mod 163), and there are infinitely many squares of this form using the above theorem.
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Old 2020-02-12, 16:50   #4
Batalov
 
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Phi(4,2^7658614+1)/2

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Quote:
Originally Posted by Batalov View Post
[

Yes infinitely many lol
@ enzocreti - Do not edit other peoples posts.
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