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 Hensel-lifting 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 :cmd: :beatdown: :uncwilly:

Bernhard

Not quite clear to me what you ask, but I think that every [URL="https://en.wikipedia.org/wiki/Pythagorean_prime"]prime of the form p = 4n+1[/URL] 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 non-trivial square, are [URL="https://oeis.org/A049532"]A049532[/URL]. /JeppeSN

[QUOTE=bhelmes;499385]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 Hensel-lifting 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 :cmd: :beatdown: :uncwilly:

Bernhard[/QUOTE]

Well there's the obvious pick n coprime to m .
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.

[QUOTE=bhelmes;499385]<snip>
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
[/QUOTE]
Assuming f(x) is a polynomial with integer coefficients, and you are only interested in integer values of x, then whether p divides f(m) only depends on m (mod p). The relevant values of m (mod p) show up in the linear factors of f(x) (mod p), i.e. f(x) viewed as a polynomial in F[sub]p[/sub][x]. Representative m-values may be taken in the interval [0, p-1].

For f(x) = x[sup]2[/sup] + 1, there will (for p == 1 (mod 4)) be two such values of m in [0, p-1]; one value will be in [0, (p-1)/2]. In either case, though, 0 < m[sup]2[/sup] + 1 < p[sup]2[/sup], so neither value is divisible by p[sup]2[/sup].

[QUOTE=Dr Sardonicus;502946]

For f(x) = x[sup]2[/sup] + 1, there will (for p == 1 (mod 4)) be two such values of m in [0, p-1]; one value will be in [0, (p-1)/2]. In either case, though, 0 < m[sup]2[/sup] + 1 < p[sup]2[/sup], so neither value is divisible by p[sup]2[/sup].[/QUOTE]


Thanks for this clear and logical answer :truck::xmastree::camping:

Bernhard