<s>[INST] Prove that for any $n \neq 4$, if for any $k < n$ we have that $a_k$ is relatively prime to $a_n$, then $a_n$ is prime 
Consider the sequence $a_n = |n(n+1)-19|$ for $n = 0,1,2,\ldots$. Prove that for any $n \neq 4$, if for any $k < n$ we have that $a_k$ is relatively prime to $a_n$, then $a_n$ is prime.

The first few terms are $19,17,13,7,1,11,23,37,53,71,91,\ldots$ and we can verify the property for each of these integers. Suppose that $a_n$ where $n \neq 4$ is relatively prime to $a_k$ for any $k < n$. Then $$\gcd(a_n,a_{n-1}) = \gcd(n(n+1)-19,(n-1)n-19) = 1.$$ Then $\gcd(n^2+n-19,n^2-n-19) = \gcd(2n,n^2-n-19) = \gcd(n,n^2-n-19) = \gcd(n,19)$ so $19 \nmid n$. Similarly $$\gcd(a_{n-1},a_{n-2}) = \gcd((n-1)n-19,(n-2)(n-1)-19).$$ Then $\gcd(n^2-n-19,n^2-3n-17) = \gcd(2n+2,n^2-3n-17) = \gcd(n+1,n^2-3n-17)$, but I didn't see how to generalize it.
 [/INST] If $a_n$ is not prime and $n > 4$, then $a_n>1$ and there exists a prime natural number $p\leq n$ such that $p\mid a_n$.  Then, $p$ divides both $a_n$ and $a_{n-p}$.  (In fact, one can show that $p=n$ implies $n=19$.)
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