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

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TopNewest\(ab=l\)

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

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

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)

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

But since \( \gcd(l+1,l)=1 \Rightarrow h=1\) – Aneesh Kundu · 2 years, 9 months ago

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Answer is 4. – Ar Agarwal · 2 years, 9 months ago

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4 is the answer – Gaurav Singh · 1 year, 10 months ago

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The only solution which is a natural number is 4. – Sahil Nare · 2 years, 2 months ago

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