# Pre-RMO 2014/9

Natural numbers $$k, l, p$$ and $$q$$ are such that if $$a$$ and $$b$$ are the roots of $$x^2 -kx + l = 0$$ then $$a + \frac{1}{b}$$ and $$b + \frac{1}{a}$$ are the roots of $$x^2 -px +q = 0$$. What is the sum of all possible values of $$q$$?

This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
3 years, 9 months ago

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$$ab=l$$

$$\frac{(l+1)^{2}}{l}=q$$

$$\Rightarrow (l+1)^{2}=lq$$

Since $$\gcd(l+1,l)=1$$ (Its also true for $$(l+1)^{2}$$)

$$(l+1)^{2}=q$$

$$l=1$$

$$\Rightarrow q=\boxed{4}$$ (q can't be negative)

Note

Even if u take $$q|(l+1)^{2} \Rightarrow (l+1)^{2}=qh$$ or $$h=l$$

But since $$\gcd(l+1,l)=1 \Rightarrow h=1$$

- 3 years, 9 months ago

- 3 years, 9 months ago

- 2 years, 10 months ago

The only solution which is a natural number is 4.

- 3 years, 2 months ago