Taking any $a$ coprime to $n$, we find $b \equiv -a^{-1} \mod n$ such that $ab\equiv -1\mod n$, thus $a+b \equiv 0 \mod n$, giving $a^2 \equiv 1 \mod n$. We just used $a$ coprime to $n$, thus we conclude like in http://www.problem-solving.be/pen/viewtopic.php?t=150 .

ZetaX wrote: giving $a^2 \equiv 1 \mod n$. How do you do this? I see $(a+b)^2-2ab=a^2+b^2\equiv 2\pmod n$, but why do $a$ and $b$ have the same $\mod n$ remainder?

We set $b=-a^{-1}$ and get $a-a^{-1} \equiv 0 \mod n$. Multiplication with $a$ gives $a^2-1 \equiv 0 \mod n$. Or with your identity: $a \equiv -b \mod n$ by the condition, giving $2a^2 \equiv a^2+b^2 \equiv 2 \mod n$.

Peter wrote: Determine all positive integers $n$ such that $ xy+1 \equiv 0 \; \pmod{n} $ implies that $ x+y \equiv 0 \; \pmod{n}$. In order to fill the ??? in the source: This is part (b) of AMM problem S 9 [1979, 306] by M. S. Klamkin and A. Liu. Reference from JSTOR: Solutions of Problems Dedicated to Emory P. Starke S9 M. S. Klamkin; A. Liu; Arnold Adelberg; Jeffrey M. Cohen The American Mathematical Monthly, Vol. 87, No. 6. (Jun. - Jul., 1980), p. 488. For the sake of completeness, part (a) states that: Determine all positive integers $n$ such that $\gcd\left(x,n\right)=1$ implies that $x^2\equiv 1\pmod n$. Darij

Would this be a generalization of A.18?