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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 5 cases: 1) no solution 2) one solution 3) two solutions 4) three solutions 5) four solutions

$${depending}\:{on}\:{the}\:{values}\:{of}\:{a}\:{and}\:{b}, \\ $$$${there}\:{are}\:\mathrm{5}\:{cases}: \\ $$$$\left.\mathrm{1}\right)\:{no}\:{solution} \\ $$$$\left.\mathrm{2}\right)\:{one}\:{solution} \\ $$$$\left.\mathrm{3}\right)\:{two}\:{solutions} \\ $$$$\left.\mathrm{4}\right)\:{three}\:{solutions} \\ $$$$\left.\mathrm{5}\right)\:{four}\:{solutions} \\ $$

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 ∈N

$${thank}\:{you}\:{very}\:{much}\:{for}\:{your}\:{answer}. \\ $$$${a}\:{and}\:{b}\:,{are}\:\in\mathbb{N} \\ $$

Answered by sma3l2996 last updated on 01/Apr/17

x=a−y^2 (a−y^2 )^2 +y^2 =b a^2 +y^4 −2ay^2 +y^2 =b y^4 −(2a−1)y^2 =b−a^2 y^4 −(((2a−1)×2)/2)y^2 +(((2a−1)/2))^2 =b−a^2 +(((2a−1)/2))^2 (y^2 −((2a−1)/2))^2 =b−a^2 +a^2 −a+(1/4)=((4(b−a)+1)/4) y^2 =+_− ((√(4(b−a)+1))/2)+((2a−1)/2) y=+_− (√((+_− (√(4(b−a)+1))+2a−1)/2)) x=a−y^2 =+_− (((√(4(b−a)+1))+2a−1)/2)+a

$${x}={a}−{y}^{\mathrm{2}} \\ $$$$\left({a}−{y}^{\mathrm{2}} \right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={b} \\ $$$${a}^{\mathrm{2}} +{y}^{\mathrm{4}} −\mathrm{2}{ay}^{\mathrm{2}} +{y}^{\mathrm{2}} ={b} \\ $$$${y}^{\mathrm{4}} −\left(\mathrm{2}{a}−\mathrm{1}\right){y}^{\mathrm{2}} ={b}−{a}^{\mathrm{2}} \\ $$$${y}^{\mathrm{4}} −\frac{\left(\mathrm{2}{a}−\mathrm{1}\right)×\mathrm{2}}{\mathrm{2}}{y}^{\mathrm{2}} +\left(\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={b}−{a}^{\mathrm{2}} +\left(\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\left({y}^{\mathrm{2}} −\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={b}−{a}^{\mathrm{2}} +{a}^{\mathrm{2}} −{a}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{4}\left({b}−{a}\right)+\mathrm{1}}{\mathrm{4}} \\ $$$${y}^{\mathrm{2}} =\underset{−} {+}\frac{\sqrt{\mathrm{4}\left({b}−{a}\right)+\mathrm{1}}}{\mathrm{2}}+\frac{\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}} \\ $$$${y}=\underset{−} {+}\sqrt{\frac{\underset{−} {+}\sqrt{\mathrm{4}\left({b}−{a}\right)+\mathrm{1}}+\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}} \\ $$$${x}={a}−{y}^{\mathrm{2}} =\underset{−} {+}\frac{\sqrt{\mathrm{4}\left({b}−{a}\right)+\mathrm{1}}+\mathrm{2}{a}−\mathrm{1}}{\mathrm{2}}+{a} \\ $$

Commented by mrW1 last updated on 01/Apr/17

line 2 should be: (a−y^2 )^2 +y=b

$${line}\:\mathrm{2}\:{should}\:{be}: \\ $$$$\left({a}−{y}^{\mathrm{2}} \right)^{\mathrm{2}} +{y}={b} \\ $$

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 mrW1, pionted:ther is a little mistake in line #2.

$${thank}\:{you}\:{for}\:{answer}.{but}\:{as}\:{mrW}\mathrm{1}, \\ $$$${pionted}:{ther}\:{is}\:{a}\:{little}\:{mistake}\:{in}\: \\ $$$${line}\:#\mathrm{2}. \\ $$

Commented by sma3l2996 last updated on 01/Apr/17

yes you alright

$${yes}\:{you}\:{alright}\: \\ $$

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