20200210, 16:03  #1 
Mar 2018
527_{10} Posts 
51+163k
numbers of the form 51+163k
conjecture 51+163k is a square only for k=10 
20200210, 17:04  #2 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×3×5×311 Posts 
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 ... ... Last fiddled with by enzocreti on 20200210 at 17:34 
20200210, 23:39  #3 
Jan 2017
2^{2}×7 Posts 
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 wellknown 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. 
20200212, 16:50  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·3·5·311 Posts 
