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The smallest possible natural number $n$, for which the equation $x^{2}-n x+2014=0$ has integral roots, is

I know the discriminant will be a perfect square, but I am struck on equation of discriminant.

Piesquareisg
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  • You can search up how to calculate the discriminant. Have you tried that yet? – Toby Mak Jun 14 '20 at 05:12
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    For a quadratic $f(x)=ax^2 + bx + c$, the discriminant is $b^2-4ac$, which in your case is $n^2-4\cdot 2014$. – K.defaoite Jun 14 '20 at 05:13
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    You want $n^2-2014\cdot4=m^2$ for some $m$. This might not be the fastest way, but you could use difference of squares (since there are only a few factors of 8056) and pick the smallest $n$. – boink Jun 14 '20 at 05:14
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    You actually have it lucky in this problem since $2014$ only has $8$ divisors. Note that the sum of the two roots is $n$ by Vieta's formulas. – Toby Mak Jun 14 '20 at 05:16
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    Note that $(a+b)^2=(a-b)^2+4ab$. If you fix the product, you should be able to see that the sum of the roots is least when the difference of the roots is least. – Mark Bennet Jun 14 '20 at 06:02

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The prime factorisation of $2014$ is $2 \times 19 \times 53$. $2$ is obviously a factor, so you need to find the factors of $1007$ by hand. To do this, you only need to check the primes less than $\sqrt{1007} \approx 31.7$ of which there aren't many.

This gives the factor pairs $(1, 2014); (2, 1007); (19, 106)$ and $(38, 53)$. By Vieta's formulas, the sum of the two roots is $\frac{-b}{a}$ in $ax^2+bx+c$, or $-\frac{-n}{1} = n$ in your case. From here, you can immediately observe what the smallest value of $n$ should be.

Toby Mak
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