 Odd-order groups have no nontrivial real irreducible characters

For a real-valued irreducible character of a finite odd-order group, pairing $g$ with $g^{-1}$ gives $⟨\chi ,1_G⟩=(\chi (1)+2A)/|G|$, with $A$ an algebraic integer. If this inner product were zero, $\chi (1)$ would be even, since a rational algebraic integer is an integer. But an irreducible character degree divides the group order, so it is odd. Therefore the character is trivial.