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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.

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    $\begingroup$ This shows that there are infinitely many $n$ where $43\mid n^2+n+41$, but doesn't immediately say anything about $42$. $\endgroup$ – hmakholm left over Monica Sep 26 '18 at 4:05
  • $\begingroup$ typo...fixing it. $\endgroup$ – Idle Math Guy Sep 26 '18 at 4:06
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    $\begingroup$ 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 :) $\endgroup$ – katana_0 Sep 26 '18 at 4:20
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    $\begingroup$ 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. $\endgroup$ – Mike Sep 26 '18 at 5:11
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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: $$f(k)=m \Rightarrow m \mid f(m\cdot n + k), \forall n$$

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

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