Quadratic equation with natural number coefficients

natural numbersquadratics

Let $a,b,c $ be Natural Numbers, such that roots of the equation $ax^2-bx+c=0$ are distinct and both lie in the interval

  1. (0,1)
  2. (1,2)
  3. (2,3)

(Brackets signify open interval, roots are $IN BETWEEN $ the numbers in each part.)

Find minimum possible value of $a, b, c.$

On my part, I solved for part 1, i.e. for distinct roots between (0,1). But for the next two parts, the things are getting a too bit messy.

While it may have similarity in question for given part 1 in stack exchange, there is no generalized method so that we can solve for other such intervals.

So please help, I am new to stack exchange.

For part 3, I tried by taking $0<m-2, 3-m, n-2, 3-n <1$ where m, n are the roots of the equation, and then using A. M. – G. M method, but i failed.

Please check another question of this type, But please don't provide with such answers as given in the link, as this is a question of an entrance exam, to be solved by hand, and not wolfram mathematica.

Consider the quadratic equation $ax^2-bx+c=0, a,b,c \in N. $ If the given equation has two distinct real root…

Please also tell me whether such numbers correspond to any famous known series.

Many are telling that there can be no such coefficients, then for part 1, please check $5x^2-5x+1=0$. It has it's roots between 0 and 1, and the coefficients are natural numbers to be sure, namely $a=5,b=5,c=1$.

Best Answer

So multiply through by $4a$ to obtain $$4a^2x^2-4abx+4ac=(2ax-b)^2+4ac-b^2$$

Now with $a$ positive the quadratic has a minimum value at $x=\cfrac b{2a}$ and the value must be negative at this point so that $4ac-b^2\lt 0$. You also need $\cfrac b{2a}$ in the relevant range of values for $x$, and if the range is the interval $(p,q)$ you need the values at $p$ and $q$ to be positive.

The question is then about controlling for these conditions whilst keeping $a,b,c$ small.

So let's try it for the interval $(2,3)$

We need $4a-2b+c\gt 0$ and $9a-3b+c\gt 0$ from the conditions at the endpoints.

We also need $2\lt \frac b{2a}\lt 3$ ie $4a\lt b\lt 6a$

The smallest possible value of $a$ is $1$ which would make $b=5$ and we'd need $-6+c\gt 0$ ie $c\gt 6$

Now $4ac-b^2=4c-25$ must be negative and that is impossible for $c\gt 6$

So try $a=2$ with $b=9,10,11$

With $b=9$ we need$-10+c\gt 0$ and $-9+c\gt 0$ ie $c\gt 10$, and $8c-81\lt 0$ which is impossible for $c\gt 10$

With $b=10$ we get $c\gt 12$ and need $8c-100 \lt 0$ which is again impossible. But there is more margin here than with $a=1$ or with the asymmetric $b=9$.

Looking to be efficient, let's try for a symmetric solution with $a$ as small as possible and $b=5a$. We can the check other possibilities later.

The edge conditions are then $4a-10a+c\gt 0$ and $9a-15a+c\gt 0$ ie $c\gt 6a$ (or for later $4c\gt 24a$)

Then $4ac-b^2=4ac-25a^2\lt 0$ so that $25a\gt 4c\gt 24a$. So we need $a$ large enough that there is a multiple of $4$ between $24a$ (which is divisible by $4$) and $25a$. The least value of $a$ is $5$ and gives $b=25$ and $c=31$

I will leave you to do the extra checking. It is tedious going through the cases, but perfectly possible to do.

Note that (as suggested in comments) this is $5(x-2)^2-5(x-2)+1$

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