20210401, 00:02  #12  
Apr 2020
223 Posts 
Quote:
In more detail: if p is not 1 mod 5 (or 2), then 16x^5 + n has exactly one root mod p, because 5 is coprime to p1 and so x > x^5 is a bijection mod p (ie every element has a unique 5th root). If p is 1 mod 5, then either 16x^5 + n has 5 roots (with probability 1/5) or no roots (probability 4/5). So 16x^5 + n will be easier if a lot of small p = 1 mod 5 happen to give the full set of 5 roots. Your poly 1 is very lucky in that it has 5 roots mod 11, 31 and 61, whereas the first prime to which poly 2 has 5 roots is 181. In addition, poly 2 does not have roots mod 9 or 49 (since the coefficient 198345 is divisible by 3 and 7 but not 9 or 49), whereas poly 1 does. Last fiddled with by charybdis on 20210401 at 00:05 

20210401, 13:42  #13 
Jan 2012
Toronto, Canada
39_{16} Posts 
Appreciate the detailed response! That was exactly what I wanted to find out

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