Find initial identities
翻訳:Putting m=n=0 m=n=0m=n=0 翻訳: gives f(f(0))=f(f(0))+f(0) f(f(0))=f(f(0))+f(0)f(f(0))=f(f(0))+f(0) 翻訳:, so f(0)=0 f(0)=0f(0)=0 翻訳:. Putting m=0 m=0m=0 翻訳: then gives f(f(n))=f(n) f(f(n))=f(n)f(f(n))=f(n) 任意の n n n, およびequati上のbecomes f(m+f(n))=f(m)+f(n) f(m+f(n))=f(m)+f(n)f(m+f(n))=f(m)+f(n).