solve in Z :
First note that may be a better way
$\displaystyle n^2+n-2=(n+2)(n-1)=0$ this gives two solutions
$\displaystyle n=1,n=-2 \equiv 40 \mod(42)$
Note now that
$\displaystyle 42=2\cdot 3 \cdot 7$
So the only zero divisors of the ring are
$\displaystyle 2\cdot 21,14\cdot 3,6 \cdot 7$
And that there is no way to choose n to get any of the above pairs.
So the above are the only two solutions
Hi dhiab.
Unfortunately, Deadstar and TheEmptySet miss two solutions: $\displaystyle n=19,\,22.$
$\displaystyle n^2+n-1\equiv0\,\bmod42\ \implies\ 42=2\cdot3\cdot7\mid(n+2)(n-1).$
If 2, 3, 7 all divide $\displaystyle n+2$ or $\displaystyle n-1$, the solution set is $\displaystyle \{1+42k:k\in\mathbb Z\}\cup\{40+42k:k\in\mathbb Z\}.$
Otherwise …
(i) If 2, 3 divide $\displaystyle n+2$ and 7 divides $\displaystyle n-1,$ then $\displaystyle n\equiv-2\,\bmod6\equiv1\,\bmod7.$ The Chinese-remainder method (or otherwise trial and error) gives $\displaystyle n=22$ as one solution. Hence the solution set is $\displaystyle \{22+42k:k\in\mathbb Z\}.$
(ii) If 3 divides $\displaystyle n+2$ and 2, 7 divide $\displaystyle n-1,$ then $\displaystyle n\equiv-2\,\bmod3\equiv1\,\bmod14.$ In this case, $\displaystyle n=1$ but we already have this solution above.
(iii) If 2 divides $\displaystyle n+2$ and 3, 7 divide $\displaystyle n-1,$ then $\displaystyle n\equiv-2\,\bmod2\equiv1\,\bmod21,$ which gives the same solution set as in (i).
(iv) If 2, 7 divide $\displaystyle n+2$ and 3 divides $\displaystyle n-1,$ then $\displaystyle n\equiv-2\,\bmod14\equiv1\,\bmod3.$ So $\displaystyle n=40$ which we already have above.
(v) If 3, 7 divide $\displaystyle n+2$ and 2 divides $\displaystyle n-1,$ then $\displaystyle n\equiv-2\,\bmod21\equiv1\,\bmod2.$ So $\displaystyle n=19$ and we have found a new solution set.
(vi) Finally, if 7 divides $\displaystyle n+2$ and 2, 3 divide $\displaystyle n-1,$ then $\displaystyle n\equiv-2\,\bmod7\equiv1\,\bmod6,$ and again we have $\displaystyle n=19.$
Hence the complete solution set is $\displaystyle \{a+42k:k\in\mathbb Z,\,a=1,19,22,40\}.$