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Working with integers...

Define $\lambda(n,k) = n^2 -kn + 1$.

Define, for each integer $k \ge 3$, the set

$\tag 1 A_k = \{n^2-kn+1 \mid n \ge k+1 \}$

Do all these sets contain an infinite number of primes?

My work

I've been investigating $\lambda$; see, for example

$\quad$Proving (or not) that two integer valued sequences are equal ($\; \lambda(n,k) = n^2 -kn + 1\;$).

and this question naturally comes to mind.

I set $k = 1000000000$ and had python check the first $1000$ numbers in $A_{1000000000}$ and it found these prime numbers (listing $n,\lambda(n,k)$):

1000000020 20000000401
1000000074 74000005477
1000000120 120000014401
1000000200 200000040001
1000000326 326000106277
1000000356 356000126737
1000000384 384000147457
1000000410 410000168101
1000000414 414000171397
1000000554 554000306917
1000000590 590000348101
1000000666 666000443557
1000000744 744000553537
1000000776 776000602177
1000000834 834000695557
1000000854 854000729317
1000000980 980000960401
CopyPasteIt
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    This is open. The Bunayakovsky conjecture predicits infinite many prime numbers if some assumptions are satisfied. But even for $\ n^2+1\ $ , we cannot prove that infinite many primes will appear , although this almost surely will be the case. – Peter Feb 25 '20 at 14:41
  • @peter So most likely throwing in $-kn$ won't be helpful. – CopyPasteIt Feb 25 '20 at 14:43
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    To find huge primes, it won't be helpful. A suggestion to find large primes is to search for a prime of the form $k\cdot p$#$+1$ , where $p$ is a relatively large prime. The chance to get a prime then, is relatively high. Also $a^{2^n}+1$ is a good approach, where $a$ is even. Many of the largest known primes are of this form. – Peter Feb 25 '20 at 14:46

0 Answers0