Question Number 156447 by MathSh last updated on 11/Oct/21
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{functions}\:\:\mathrm{f}\::\:\mathbb{Z}\:\rightarrow\:\mathbb{R}\:\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{f}\left(\mathrm{n}+\mathrm{m}\right)=\mathrm{nf}\left(\mathrm{n}\right)+\mathrm{mf}\left(\mathrm{m}\right)+\mathrm{nm}-\mathrm{n}-\mathrm{m} \\ $$$$\forall\mathrm{n};\mathrm{m}\in\mathbb{Z} \\ $$
Answered by ghimisi last updated on 11/Oct/21
$${m}=\mathrm{1};{n}=\mathrm{0}\Rightarrow{f}\left(\mathrm{1}\right)={f}\left(\mathrm{1}\right)−\mathrm{1}\Rightarrow\mathrm{0}=−\mathrm{1} \\ $$$$???????????????? \\ $$$$ \\ $$
Commented by MathSh last updated on 11/Oct/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{wrong}.? \\ $$