Question Number 145202 by mnjuly1970 last updated on 03/Jul/21
Commented by loveineq last updated on 03/Jul/21
$$\mathrm{Wow},\:\mathrm{it}\:\mathrm{likes}\:\mathrm{a}\:\mathrm{therom}. \\ $$
Answered by ajfour last updated on 03/Jul/21
$$\left({x}+\mathrm{3}{p}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\left({x}+{p}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\left({x}−{p}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$$\left({x}+\mathrm{3}{p}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={d}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} =\mathrm{4}{x}^{\mathrm{2}} +\mathrm{12}{p}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} \\ $$$${d}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} =\mathrm{4}{x}^{\mathrm{2}} +\mathrm{12}{p}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} \\ $$$${hence} \\ $$$${a}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} ={d}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 03/Jul/21
$${thank}\:{you}\:{so}\:{much}\:{mr}\:{ajfor} \\ $$
Answered by mr W last updated on 03/Jul/21
$$\frac{{x}^{\mathrm{2}} +{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{xb}}=−\frac{{x}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{xb}} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} ={a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$${similarly} \\ $$$$\Rightarrow\mathrm{2}{x}^{\mathrm{2}} ={b}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} \:\:\:...\left({ii}\right) \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} ={b}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} ={d}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 03/Jul/21
$${thank}\:{alot}\:{mr}\:{W}\:,\:{very}\:{nice}\: \\ $$$$\:{solution}... \\ $$