Can $(2n-1)^2+2(2n-1)k$ generate primes for $n>1$?

Question

I have a function $$A=(2n-1)^2+2(2n-1)k$$ that I have proven, to myself, to generate $all$ values of $A$ for Pythagorean triples where GCD(A,B,C) is an odd square – so it includes all primitives. Let's assume my proof is valid because this is not about generating triples. In this function, $n$ is the number of a set of triples where the difference between $B$ and $C$ is the $n^{th}$ odd square and $k$ is the number of the member triple in that set.

$Set_1$ ($n=1$) includes all odd numbers $>1$ so it includes all prime numbers except $2$. My problem is that I have not been able to find prime numbers among those generated when $n>1$. Can this function generate primes for $n\ge 2$ or there something about it that will not allow and f(n,k) to generate primes when $n>1?

score 6 · Accepted Answer · answered Apr 15 '19 at 17:30

6

$(2n-1)^2+2(2n-1)k$ is never a prime when $n>1$ (assuming $k\ge 0$) because $$ (2n-1)^2+2(2n-1)k = (2n-1)(2n-1+2k) $$

answered Apr 15 '19 at 17:30

hmakholm left over Monica

286,031

Thanks. I never thought to expand, collect terms, and factor. – poetasis Apr 15 '19 at 17:53
1

@poetasis: I didn't even do that -- just noticed that each of the terms in your expression is explicitly a multiple of $2n-1$. – hmakholm left over Monica Apr 15 '19 at 18:04
Thanks anyway. It helped me help a guy with another question where I failed miserably before. – poetasis Apr 15 '19 at 18:12

Can $(2n-1)^2+2(2n-1)k$ generate primes for $n>1$?

1 Answers1