Question Number 90013 by ar247 last updated on 20/Apr/20
$$−{p}^{\mathrm{2}} +\mathrm{2027}=−{q}^{\mathrm{2}} \\ $$$${p}+{q}=? \\ $$
Commented by mr W last updated on 21/Apr/20
$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} =\mathrm{2027}\:\leftarrow\:{prime} \\ $$$$\left({p}−{q}\right)\left({p}+{q}\right)=\mathrm{1}×\mathrm{2027} \\ $$$${for}\:{p},{q}\in{N} \\ $$$${p}+{q}=\mathrm{2027} \\ $$$${p}−{q}=\mathrm{1} \\ $$$$\Rightarrow{p}=\mathrm{1014},\:{q}=\mathrm{1013} \\ $$
Commented by MJS last updated on 21/Apr/20
$$\mathrm{for}\:\mathrm{your}\:\mathrm{solution},\:{u}=\frac{\mathrm{2027}}{\mathrm{2}} \\ $$
Answered by MJS last updated on 21/Apr/20
$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} −\mathrm{2027}=\mathrm{0} \\ $$$${p}={u}−{v}\wedge{q}={u}+{v} \\ $$$$−\mathrm{4}{uv}−\mathrm{2027}=\mathrm{0} \\ $$$${v}=−\frac{\mathrm{2027}}{\mathrm{4}{u}} \\ $$$${p}={u}+\frac{\mathrm{2027}}{\mathrm{4}{u}}\wedge{q}={u}−\frac{\mathrm{2027}}{\mathrm{4}{u}} \\ $$$${p}+{q}=\mathrm{2}{u}\:\mathrm{with}\:{u}\neq\mathrm{0} \\ $$$$\Rightarrow\:{p}+{q}={z};\:{z}\in\mathbb{C}\backslash\left\{\mathrm{0}\right\} \\ $$