What is the smallest possible natural number \(n\) for which the equation \(x^2 - nx + 2014 = 0\) has integer roots?

This note is part of the set Pre-RMO 2014

What is the smallest possible natural number \(n\) for which the equation \(x^2 - nx + 2014 = 0\) has integer roots?

This note is part of the set Pre-RMO 2014

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestLet the roots of the equation be \(\alpha\) and \(\beta\).

Then \(n = \alpha + \beta\) and \(2014 = \alpha \beta\).

The possible integral pairs of \((\alpha, \beta)\) are \((1, 2014), (2, 1007), (19, 106),\) and \((38, 53)\).

Therefore, the possible values of \(n\) are \(2015, 1009, 125\) and \(91\). The minimum value is \(91\), so \(n = \boxed{91}\) – Pranshu Gaba · 2 years ago

Log in to reply

Is the answer 91? – Abhineet Nayyar · 2 years ago

Log in to reply

– Krishna Ar · 2 years ago

Yes , it is 91 :DLog in to reply

– Abhineet Nayyar · 2 years ago

Yayy!! lol:D:DLog in to reply

91 – Gaurav Singh · 1 year, 2 months ago

Log in to reply

91 – Akshat Sharda · 1 year, 2 months ago

Log in to reply

91 – Sahil Nare · 1 year, 6 months ago

Log in to reply

The discriminant of the above equation should be a perfect square for the equation to have integral roots. So the smallest value of n for which it is a perfect square is 91. – Sanchit Ahuja · 2 years ago

Log in to reply

Explain? – Shiv Kumar · 2 years ago

Log in to reply