# What is the smallest possible natural number $𝑛$ for which $x^{2}-nx+2014=0$ has integer roots?

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

My idea was, If the roots are integers, then they are the divisors of $2014$, I don't know if it's true or not.

If the roots are $\alpha, \beta$ then we have $\alpha \beta = 2014.$ As they are integers, $\alpha, \beta$ must be divisors of $2014=2\cdot19\cdot53$, giving $\left \{ \alpha, \beta \right \}=\left \{ 1,2014 \right \}$, or $\left \{ -1,-2014 \right \}$ or $\left \{ 2,1007 \right \}$ or $\left \{ -2,-1007 \right \}$ or $\left \{ 19,106 \right \}$ or $\left \{ -19,-106 \right \}$ or $\left \{ 38,53 \right \}$ or $\left \{ -38,-53 \right \}$.

Also $n=\alpha+\beta$, has the least positive value of $n$ as $38+53=91$

AND WE'RE DONE

We have for natural number $a,b$ $$ab=2014=2\cdot19\cdot53$$

and we need to minimize $n=a+b$

Check for $1,2014;$

$2,2014/2;$

$19;2014/19;$

$2\cdot19,2014/2\cdot19;$

$53,2014/53$ etc.

We have: $\triangle = b^2-4ac = n^2 - 4(2014) = k^2 \to (n-k)(n+k) = 4\times 2014$. From this we can list all possible values of $n$, and select the smallest candidate.