If $a+b\equiv0\pmod m$ and $c+d\equiv 0\pmod m$ then $ac\equiv bd\pmod m$. Demonstration.

Question

If $a+b\equiv0\pmod m$ and $c+d\equiv 0\pmod m$ then $ac\equiv bd\pmod m$ How to show?$$$$ I tried $$a+b\equiv0\pmod m\Longrightarrow m\mid a+b\\c+d\equiv0\pmod m\Longrightarrow m\mid c+d\\a+b=mk\;\;\text{and}\;\;c+d=mj\\(a+b)(c+d)=m(mjk)\\m\mid ab+ad+bc+bd$$ But from here could solve anything.

Answer 1

Hint $$ ac - bd = (a+b)c - b(c+d) $$

Answer 2

As $a\equiv -b,c\equiv-d\pmod m$

$$ac\equiv (-b)(-d)\pmod m\equiv bd$$

Answer 3

If $$a+b\equiv0\pmod m\Longrightarrow m\mid a+b\Longrightarrow m\mid(a+b)c\Longrightarrow \fbox{$ac+bc=mk$}\;\;\text{and}\\c+d\equiv0\pmod m\Longrightarrow m\mid c+d\Longrightarrow m\mid (c+d)b\Longrightarrow \fbox{$bc+bd=mj$}$$ That $$ac+bc-bc-bd=mk-mj\\ac-bd=m(k-j)\\m\mid ac-bd\Longrightarrow ac\equiv bd\pmod m\;\;\;\;\;\;\Box$$