When is f(n) = n^2+n+17 a composite number?

[MattcAnderson]

I have done similar explorations for h(n) = n^2 + n + 41
See mattanderson.fun and scroll down to 'downloads' see file 'prime poly 41.pdf
Another reference is wolfram 'Lucky Numbers of Euler'
For polynomials of the form n^2 - n + p, many prime numbers can be found when n is an integer, and especially when p is a lucky number of Euler. The six lucky numbers of Euler are 2, 3, 5, 11, 17, and 41.

I have forgotten what I did for h(n) and I have started looking at f(n) = n^2+n+17.
From a computer calculation, I find that
f(0) mod 17 is congruent to 0
f(1) mod 19 is congruent to 0 and
f(2) mod 23 is congruent to 0
I try the method of quadratic curve fit, to find a general case where f(A*x^2+B*x+C) factors algebraically.
Here, I calculated that
D = A*x^2+B*x+C
and the three data points above, calculates to
A=B=1, C=17

This is kind of a mess, and I was hoping someone else could take a look at it.

My question is, for what values of A,B,and C does f(A*x^2+B*x+C) factor algebraically?

Regards,
Matt

[R.D. Silverman]

Please move to misc.math

[MattcAnderson]

I sort of answered my own question.
I have found that if f(n) = n^2+n+17
and n= t^2+16
then substituting
f(t^2+16) = (t^2+16)^2 + (t^2+16) + 17
and factoring the right hand side gives
f(t^2+16) = (t^2+t+17)*(t^2-t+17)
These are some of the cases when f(n) is a composite number.
We use t = 0,1,2 ...
This corresponds with the data point in one of the parabolas of the graph from before.

Have a nice day.

Matt

[bhelmes]

The answer when f(n)=ax²+bx+c is a composite value
is a solution of the chinese remainder theorem,
the set of the congruences has something to do with the primes p | f(n) for n=0...n0
and perhaps some senseful thoughts and an algorithm especially for you and for that is
http://devalco.de/basic_polynomials/...p?a=1&b=1&c=17

Enjoy the algorithm (a sieving construction similar to the sieve of Eratosthenes)

[MattcAnderson]

Thank you bhelmes for your input.
I assume the system is linear.
I must solve a system of 3 equations and 3 unknowns to find a,b,and c.

It looks like I am not the only one who has put effort into this polynomial.
f(n) = n^2 + n + 17
I found a second family of factors
From my Maple graph in post 1, I have the set of 3 equations and 3 unknowns.
t y
0 33
-1 34
1 36
Using Maple CurveFitting, I find that
y = 2*t^2 + t + 33.
Let n=y and then factor the resulting quartic polynomial
f(2*t^2 + t + 33) = (4*t^2+67)*(t^2+t+17).
t is in the integers
Here, f(y) is a set of cases when f(y) is a composite number.
I am happy with this result.
Have a nice day.