Existence of an Euler witness for every odd composite integer

Every odd composite integer has an Euler witness. For a nonsquarefree integer, a base congruent to $1+p$ modulo a repeated prime factor $p^2$ fails the Euler congruence by the binomial theorem. For a squarefree integer, use the Chinese remainder theorem to choose a base that is a quadratic nonresidue modulo one prime factor and $1$ modulo another. Its Jacobi symbol is $-1$, while its power is $1$ modulo the second factor.