Not sure how to approach this problem. First idea was proof by contradiction. Assume a divides n^2 + 10 and proceed from there. I couldn't reach a substantial conclusion from this approach.

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    $\begingroup$ This would appear to be a question about Quadratic Reciprocity. $\endgroup$ – lulu Apr 21 at 14:20
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    $\begingroup$ @lulu, I wonder whether there is an elementary solution that does not use QR. $\endgroup$ – lhf Apr 21 at 14:51
  • $\begingroup$ @lhf Yes, I'd like to see that too $\endgroup$ – Vincent Apr 22 at 12:37

$a\mid n^2+10$ for some $n$ is equivalent to $-10$ being a quadratic residue modulo $a$. Now $$\left(\frac{-10}a\right)=\left(\frac{-1}a\right)\left(\frac{2}a\right)\left(\frac{5}a\right)$$ and if $a\equiv29\bmod40$ the following hold true (check e.g. Wikipedia for confirmation): $$\left(\frac{-1}a\right)=+1\qquad\left(\frac2a\right)=-1\qquad\left(\frac5a\right)=+1$$ Thus $\left(\frac{-10}a\right)=-1$, i.e. $-10$ is not a quadratic residue modulo $a$, and the question claim follows.

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