# Question-11813

Question Number 11813 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Apr/17
Commented by mrW1 last updated on 01/Apr/17
$${depending}\:{on}\:{the}\:{values}\:{of}\:{a}\:{and}\:{b}, \\$$$${there}\:{are}\:\mathrm{5}\:{cases}: \\$$$$\left.\mathrm{1}\right)\:{no}\:{solution} \\$$$$\left.\mathrm{2}\right)\:{one}\:{solution} \\$$$$\left.\mathrm{3}\right)\:{two}\:{solutions} \\$$$$\left.\mathrm{4}\right)\:{three}\:{solutions} \\$$$$\left.\mathrm{5}\right)\:{four}\:{solutions} \\$$
Commented by mrW1 last updated on 01/Apr/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Apr/17
$${thank}\:{you}\:{very}\:{much}\:{for}\:{your}\:{answer}. \\$$$${a}\:{and}\:{b}\:,{are}\:\in\mathbb{N} \\$$
Answered by sma3l2996 last updated on 01/Apr/17
$${x}={a}−{y}^{\mathrm{2}} \\$$$$\left({a}−{y}^{\mathrm{2}} \right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={b} \\$$$${a}^{\mathrm{2}} +{y}^{\mathrm{4}} −\mathrm{2}{ay}^{\mathrm{2}} +{y}^{\mathrm{2}} ={b} \\$$$${y}^{\mathrm{4}} −\left(\mathrm{2}{a}−\mathrm{1}\right){y}^{\mathrm{2}} ={b}−{a}^{\mathrm{2}} \\$$$${y}^{\mathrm{4}} −\frac{\left(\mathrm{2}{a}−\mathrm{1}\right)×\mathrm{2}}{\mathrm{2}}{y}^{\mathrm{2}} +\left(\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={b}−{a}^{\mathrm{2}} +\left(\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\$$$$\left({y}^{\mathrm{2}} −\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={b}−{a}^{\mathrm{2}} +{a}^{\mathrm{2}} −{a}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{4}\left({b}−{a}\right)+\mathrm{1}}{\mathrm{4}} \\$$$${y}^{\mathrm{2}} =\underset{−} {+}\frac{\sqrt{\mathrm{4}\left({b}−{a}\right)+\mathrm{1}}}{\mathrm{2}}+\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}} \\$$$${y}=\underset{−} {+}\sqrt{\frac{\underset{−} {+}\sqrt{\mathrm{4}\left({b}−{a}\right)+\mathrm{1}}+\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}} \\$$$${x}={a}−{y}^{\mathrm{2}} =\underset{−} {+}\frac{\sqrt{\mathrm{4}\left({b}−{a}\right)+\mathrm{1}}+\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}+{a} \\$$
Commented by mrW1 last updated on 01/Apr/17
$${line}\:\mathrm{2}\:{should}\:{be}: \\$$$$\left({a}−{y}^{\mathrm{2}} \right)^{\mathrm{2}} +{y}={b} \\$$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/Apr/17
$${thank}\:{you}\:{for}\:{answer}.{but}\:{as}\:{mrW}\mathrm{1}, \\$$$${pionted}:{ther}\:{is}\:{a}\:{little}\:{mistake}\:{in}\: \\$$$${line}\:#\mathrm{2}. \\$$
Commented by sma3l2996 last updated on 01/Apr/17
$${yes}\:{you}\:{alright}\: \\$$