<s>[INST] If $-1$is a root for $ax^2+bx−3$, find $a^2+b^2$ 
Given: -1 is a root for $ax^2+bx-3$, with $a,b$ being positive primes, $x\in \Bbb R$.
Find: the numeric value for $a^2+b^2$.

Background: question asked in an entrance exam (Colégio Militar 2005).
My attempt: the other root is $3/a$ and by substitution we can easily find that $$a-b=3\ \ \text{or}\ \ a^2+b^2-2ab=9.$$
I got stuck at this point... how to get the value for $ab$? Hints please.
 [/INST] Starting from $a-b=3$, which is odd, this implies the positive primes cannot be  both odd. So one of them is $2$, the other is an odd prime. As $a-b>0$, it's $b$ which is equal to $2$, so $a=3+b=5$, and $a^2+b^2=29$.
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