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

Question

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.

This shows that there are infinitely many $n$ where $43\mid n^2+n+41$, but doesn't immediately say anything about $42$. — hmakholm left over Monica, Sep 26 '18 at 04:05
Yes, your proof is correct. Note that for any $f(x) \in \mathbb{Z}[x]$ and $a \neq b \in \mathbb{Z}$, we have $a-b | f(a) - f(b)$, and setting $a = 43k+1, b = 1$ yields the conclusion :) — katana_0, Sep 26 '18 at 04:20
Looks good. Also note that $n^2+n+41\equiv n^2+n-2\pmod{43}$. $n^2+n-2=(n+2)(n-1)$, so $n=43k-2$ should be a solution too. — Mike, Sep 26 '18 at 05:11

score 1 · Accepted Answer · answered Sep 26 '18 at 08:40

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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: $$f(k)=m \Rightarrow m \mid f(m\cdot n + k), \forall n$$

answered Sep 26 '18 at 08:40

rtybase

16,907

Off topic, but, Is Hardy's book really good? i.e. would you recommend it to someone at, say, the advanced undergrad level? – Idle Fool Sep 26 '18 at 13:17
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@IdleMathGuy it was updated several times, I personally like it. Check the pdf version ... Or content/index with Amazon preview. – rtybase Sep 26 '18 at 13:56

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

1 Answers1