Find all real numbers $a$, $b$ and $c$ that satisfy the following system of equations: $$\begin{cases} ab-c = 3 \\ a+bc = 4 \\ a^2+c^2 = 5\end{cases}$$
Problem
Source: Kosovo Math Olympiad 2025, Grade 10, Problem 1
Tags: algebra, system of equations, Kosovo, national olympiad, nice
RagvaloD
17.11.2024 23:39
$(ab-c)^2+(a+bc)^2 =a^2b^2+c^2+a^2+b^2c^2=25=(a^2+c^2)^2$ so $b^2+1=a^2+c^2=5 \to b =\pm 2$ Rest is easy $(a,b,c)=(2,2,1),(-\frac{2}{5},-2, -\frac{11}{5})$
hongxiaobbb2
17.11.2024 23:45
RagvaloD wrote: $(ab-c)^2+(a+bc)^2 =a^2b^2+c^2+a^2+b^2c^2=25=(a^2+c^2)^2$ so $b^2+1=a^2+c^2=5 \to b =\pm 2$ Rest is easy $(a,b,c)=(2,2,1),(-\frac{2}{5},-2, -\frac{11}{5})$
AshAuktober
18.11.2024 10:02
Note that $5(b^2 + 1) = (b^2+1)(a^2+c^2) = (ab-c)^2 + (a+bc)^2 = 25 \implies b = \pm 2$. From here manually check to find $a$ and $c$.
DensSv
28.11.2024 23:57
Notice that $(3,4,5)$ is a pithagorean triplet, thus $(ab-c)^{2}+(a+bc)^{2}=(a^2+c^2)^{2}$, upon expanding the right handside we obtain that $a^{2}b^{2}+a^{2}+c^{2}+b^{2}c^{2}=(a^2+c^2)^{2}\Rightarrow (b^{2}+1)(a^2+c^2)=(a^2+c^2)^{2}\Rightarrow b^{2}+1=a^2+c^2=5$ hence $b$ is either equal to $2$ or $-2$. After substituting $b$ in the first two equations we obtain the following solutions $(a,b,c)=(2,2,1),(-\frac{2}{5},-2, -\frac{11}{5})$
lksb
29.11.2024 00:24
$(ab-c)^2+(a+bc)^2=25\implies \boxed{b=\pm2}$ With this, it's easy from the first 2 equations that the only solutions are $\boxed{(a,b,c)=\{(2,2,1),(-\frac{2}{5}, -2, -\frac{11}{5})\}}$