p² | (n²+1)

[bhelmes]

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

Bernhard

[JeppeSN]

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 non-trivial square, are A049532. /JeppeSN

[science_man_88]

Originally Posted by bhelmes

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

Bernhard

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.

[Dr Sardonicus]

Originally Posted by bhelmes

<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

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_p[x]. Representative m-values may be taken in the interval [0, p-1].

For f(x) = x² + 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 < m² + 1 < p², so neither value is divisible by p².

[bhelmes]

Originally Posted by Dr Sardonicus

For f(x) = x² + 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 < m² + 1 < p², so neither value is divisible by p².

Thanks for this clear and logical answer

Bernhard