$n^2-79n+1601$ always a prime?

Question

I am struggling with proving or disproving this:

$n^2-79n+1601$ is a prime for all natural numbers $n$ (except multiples of $1601$).

This somehow has a relation to Stanislaw Ulam spiral.

What would be a good approach?

Answer 1

Let $f(n)$ be our polynomial. Note that $f(1)=1523$. I won't bother to check whether this is prime.

For $f(1+1523)$ is divisible by $1523$, and a lot bigger, so it must be composite.

Remark: An argument that fundamentally is the one of the answer shows that if $f(n)$ is a non-constant polynomial with integer coefficients, then $f(n)$ cannot be always prime.

Answer 2

As I thought, taking $m = x + 39,$ this is $$ x^2 - x + 41. $$ This sets the record for consecutive primes, $x=1,2,3,...,40.$

If $x = t+1$ we get $t^2 + t + 41.$ No important difference, This way it is primes $t=0,1,...,39,$ but divisible by $41$ for $t=40,41.$ Also a pretty good performance for $t > 40$ and $t \neq -1,0 \pmod {41},$ but nobody knows whether it (or any other one-variable polynomial) represents infinitely many primes.

See Is the notorious $n^2 + n + 41$ prime generator the last of its type?