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Here is a problem from this year’s OMO:

Let $S$ be the set of all pairs $(a,b)$ of real numbers satisfying $1+a+a^2+a^3 = b^2(1+3a)$ and $1+2a+3a^2 = b^2 - \frac{5}{b}$. Find $A+B+C$, where $$ A = \prod_{(a,b) \in S} a , \quad B = \prod_{(a,b) \in S} b , \quad \text{and} \quad C = \sum_{(a,b) \in S} ab. $$

There are no solutions on the AoPS thread for this problem. I tried Vieta’s formulas and did not make any useful findings. Are there any clever solutions that do not use Vieta’s (or do, although I think this is very unlikely)? Thank you in advance!

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You can use Vieta's but not right away! Here is my solution, which also is the official one. First, note that $(a,b)=(1,-1)$ is an easy solution. Then, let $z=a-bi$. Multiply the second equation by $-bi$ and multiply by the first and we find $z^3+z^2+z=-1+5i$. Since $(1,-1)$ is a solution, this motivates us to factor out $ z - \left( 1 + i \right) $. We find $$ z^2 + (2+i) z + (2+3i) = 0. $$Now, let $(r,s)$ and $(t,u)$ be the other solutions (think about why we know there are only three). Then, by Vieta's: $$ \begin {eqnarray*} r + s &=& -2, \\ s + t &=& 1, \\ rs - tu &=& 2, \\ ru + st &=& -3. \end {eqnarray*} $$Now, it should be easy to find $rs+tu=1$, so $C=0$ and $A+B=2$, so our answer is $\boxed{2}$.

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