Completion of a Latin square from its diagonal

If $Δ 𝑖 = 𝑀 𝑖 𝑖$ is fixed for $𝑖 = 1, \dots, 𝑛$ , there exists an $𝑛 \times 𝑛$ Latin square $(𝑀 𝑖 𝑗)$ except if the diagonal $Δ 𝑖$ consists of exactly $𝑛 - 1$ occurances of one symbol.¹ comb