Pre-RMO 2014/9

Natural numbers k,l,pk, l, p and qq are such that if aa and bb are the roots of x2kx+l=0x^2 -kx + l = 0 then a+1ba + \frac{1}{b} and b+1ab + \frac{1}{a} are the roots of x2px+q=0x^2 -px +q = 0. What is the sum of all possible values of qq?


This note is part of the set Pre-RMO 2014

#Algebra #Pre-RMO

Note by Pranshu Gaba
6 years, 8 months ago

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Comments

Answer is 4.

Ar Agarwal - 6 years, 8 months ago

ab=lab=l

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

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

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

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

l=1l=1

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

Note

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

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

Aneesh Kundu - 6 years, 8 months ago

The only solution which is a natural number is 4.

Sahil Nare - 6 years, 1 month ago

4 is the answer

gaurav singh - 5 years, 9 months ago
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