20181102, 22:09  #1 
Mar 2016
5^{2}×11 Posts 
p²  (n²+1)
A peaceful and pleasant night for all members,
Regarding the irreducible polynomial f(n)=n²+1 If i have a prime p f(n) then i know that i could calculate p²  f(m) with the Hensellifting with m>n. If i have p²  f(m) how could i calculate p  f(n) with 1<n<m Or in other words could there be a quadrat of a prime as a divisor of the function without the earlier appearance of the prime. Perhaps someone knows an easy prove, would be nice to know it Greetings from Landaus problem Bernhard Last fiddled with by bhelmes on 20181102 at 22:12 
20181216, 00:16  #2 
"Jeppe"
Jan 2016
Denmark
2^{5}·5 Posts 
Not quite clear to me what you ask, but I think that every prime of the form p = 4n+1 will satisfy that p^2 divides numbers of the form n^2 + 1 (and other primes p will not). On the other hand, the n values such that n^2 + 1 is divisible by a nontrivial square, are A049532. /JeppeSN

20181216, 01:08  #3  
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
Quote:
There's also that n^4+2n^2+1 will divide by p^2 so gcd of these polynomials will also have the p^2 factor. Last fiddled with by science_man_88 on 20181216 at 01:08 

20181216, 01:50  #4  
Feb 2017
Nowhere
3×7×13^{2} Posts 
Quote:
For f(x) = x^{2} + 1, there will (for p == 1 (mod 4)) be two such values of m in [0, p1]; one value will be in [0, (p1)/2]. In either case, though, 0 < m^{2} + 1 < p^{2}, so neither value is divisible by p^{2}. 

20181217, 16:54  #5  
Mar 2016
100010011_{2} Posts 
Quote:
Thanks for this clear and logical answer Bernhard 
