a number whose additive inverse and multiplicative inverse are same in modular arithmetic?
Zero: Regardless of the base, m: 0(mod m) + 0(mod m) 0(mod m) and: 0(mod m) X 0(mod m) 0(mod m)
If a mod m has a multiplicative inverse, a^-1 then aa^-1 1 mod m a is relatively prime to m. Clearly the additive inverse of a mod m -a mod m. If -a a^-1 mod m then -a^2 aa^-1 1 mod m a^2 -1 mod m. -1 is a quadratic residue mod m. If for example m a prime number p then a^2 -1 mod p (-1)^{(p-1)/2} (a^2)^{(p-1)/2} a^(p-1) 1 mod m (this last is by Fermat's little theorem). Then (-1)^{(p-1)/2} 1 (p-1)/2 is even p 1 mod 4. On the other hand if p 1 mod 4, then let a 2*3*.*{(p-1)/2}, and b (p-2)*.*{p - (p-1)/2}. Then the inverse of any factor i in a is contained in either a or b, and the inverse of any factor j in b is contained in a or b ab 1 mod p. But (p-1)*(p-3).*(p-(p-1)/2) (-1)*(-2)*.*(-(p-1)/2) mod p (-1)^{(p-3)/2}a mod p -a mod p -a^2 1 mod p a^2 -1 mod p. So x^2 -1 mod p is always solvable if and only if p 1 mod 4 when p is prime. From then on you see that x^2 -1 mod p^n is always solvable using taylor series (Hensel's Lemma). Using the Chinese remainder theorem if m p(1)^e(1).p(r)^e(r) and p(i) 1 mod 4 for all i then the equation x^2 -1 mod m is always solvable. As an example let m 13, then 5^2 25 -1 mod 13, and 5^-1 8 mod 13, since 5*8 40 1 mod 13, and -5 mod 13 8 mod 13. So 5^-1 - 5 mod 13.
Yes, it could. Find another safer way to get enough light to read by.
