Sistem simplu (own)

[Stefan Tudose]

Sa se determine $a,b\in \Bbb{R}$ cu proprietatea ca $a^2=b+1,\ b^2=4a+1$

[Pricope Tidor-Vlad]

Din prima relatie obtinem:$b=a^2-1$ si inlocuind in a doua:$(a^2-1)^2=4a+1 \Leftrightarrow a^4-2a^2+1=4a+1 \Leftrightarrow a^4-2a^2=4a (1)$.Daca $a=0 \Rightarrow b=-1$,convine.Daca $a \neq 0$, (1) devine: a^3-2a=4 <=> a^3-8-2a+4=0 <=>(a-2)(a^2+2a+4)-2(a-2)=0 <=> (a-2)(a^2+2a+2)=0 de unde a=2 deoarece $a^2+2a+2 > 0$ si b=3.

[Stefan Tudose]

Solutia mea:

$4a^2+(4a+1)=4a^2+b^2=4(b+1)+b^2\Leftrightarrow (2a+1)^2=(b+2)^2$

Cazul $2a+1=-(b+2)$ nu convine (se demonstreaza usor), ramanand $2a+1=b+2$ care genereaza solutia $a=2,\ b=3$

[Virgil Nicula]

PP. Sa se determine $\{a,b\}\subset\Bbb{R}$ cu proprietatea ca $a^2=b+1\ ,\ b^2=4a+1$.

Demonstratie. Se observa ca $\boxed{a=0\iff b=-1}$ . Presupunem ca $a\ne 0$ , adica $b\ne -1$ . Asadar, $\left\{\begin{array}{c} a^2=b+1\\\\ b^2=4a+1\end{array}\right\|\iff$

$\left\{\begin{array}{c} a^2=b+1\\\\ 4a=b^2-1\end{array}\right\|\implies$ $\frac {a^2}{4a}=\frac {b+1}{b^2-1}\iff$ $\boxed{a=\frac 4{b-1}}\ (*)\implies$$\left(\frac 4{b-1}\right)^2=b+1\iff$ $(b-1)\left(b^2-1\right)=16\iff$

$b^3-b^2-b-15=0\iff (b-3)\left(b^2+2b+5\right)=0\implies$ $\boxed{b=3\ \wedge\ a=2}$ deoarece $b^2+2b+5=(b+1)^2+4>0$ .