Show that there are infinitely many integers $n$ such that $43 \mid(n^2+n+41)$.

Question

Show that there are infinitely many integers $n$ such that $43 \mid(n^2+n+41)$.

Assume $f(n) = (n^2+n+41)$. Then $f(1)=1^2+1+41 = 43$. So $f(1) \equiv 0 \pmod{43}$. If $n \equiv 1 \pmod{43}$ then $f(n)=0 \pmod{43}$.

Is this proof correct? I'm just starting to learn some number theory and I'm not as confident in this as, say, real analysis or linear algebra.

Any help is appreciated.

Answer 1

Yes, it looks correct. Just for the record, this technique is covered in Hardy's "An Introduction To The Theory Of Numbers" (more details here), if $f(x)$ is a polynomial then: $$f(k)=m \Rightarrow m \mid f(m\cdot n + k), \forall n$$