Distinct roots of $ax^2-bx+c=0$ in $(0,1)$, where $a,b,c\in \mathbb{Z}^+$

Solution 1:

The brute force approach works out fairly quickly in this case.

• $f(0) \gt 0$ $\implies$ $c \gt 0$ and $f(1) \gt 0$ $\implies$ $b \lt a + c$

• the product of the roots must be in $(0,1)$ $\implies$ $c < a$

• the discriminant $\Delta \gt 0$ $\implies$ $b^2 \gt 4 ac$

It follows that $c \ge 1$, $a \ge c+1$ and $b \in (\sqrt{4ac}, a+c)$. Trying the first few values starting at the lowest possible $c=1$ and $a=2$ gives:

$$ \begin{align} & c = 1 & a = 2 \quad \implies \quad & b \in (\sqrt{8},3) & = & \;\emptyset \\ & & a = 3 \quad \implies \quad & b \in (\sqrt{12},4) & = & \;\emptyset \\ & & a = 4 \quad \implies \quad & b \in (\sqrt{16},5) & = & \;\emptyset \\ & & a = 5 \quad \implies \quad & b \in (\sqrt{20},6) & = & \;\{\;5\;\} \end {align} $$

The latter gives the solution $5 x^2 - 5x +1$ which can be easily verified to have both roots in $(0,1)$.

$$ \begin{align} & c = 2 & a = 3 \quad \implies \quad & b \in (\sqrt{24},5) & = & \;\emptyset \\ & & a = 4 \quad \implies \quad & b \in (\sqrt{32},6) & = & \;\emptyset \end {align} $$ $$ \begin{align} & c = 3 & a = 4 \quad \implies \quad & b \in (\sqrt{48},7) & = & \;\emptyset \end {align} $$

Solution 2:

After a thinking a lot on the question I was able to come up with a good enough solution.

Let $f(x)=ax^2-bx+c$. Let the roots of the equation $f(x)=0$ be $\alpha$ and $\beta$. Then, $\alpha,\beta\in(0,1)$.

As the roots of the equation $f(x)=0$ are real and distinct, then $b^2\gt4ac$.Also, due to the roots lying in $(0,1)$, hence $f(0)\gt0$, $f(1)\gt0$ and $f(0)f(1)\gt0$

Now, $$f(0)=c\ge1\tag{1}$$ $$f(1)=a-b+c\ge1\tag{2}$$ $$\therefore \qquad f(0)\cdot f(1)\ge1\qquad\qquad\tag{3}$$

If $\alpha$ and $\beta$ are the zeroes of $f(x)$, then $f(x)=a(x-\alpha)(x-\beta)$ $$\therefore f(0)\cdot f(1)=a\alpha(1-\alpha)a\beta(1-\beta)\ge 1$$

Using A.M-G.M inequality we get, $$\dfrac{b+(1-b)}{2}\ge\sqrt{b(1-b)}\implies b(1-b)\le\dfrac{1}{4}$$ equality occurs when $b=\dfrac{1}{2}$ $$\therefore \qquad\qquad\alpha(1-\alpha)\le\dfrac{1}{4}\qquad\qquad \beta(1-\beta)\le\dfrac{1}{4}$$

In both the cases equality occurs when $\alpha=\beta=\dfrac{1}{2}$, but as $\alpha\neq\beta$

So, $$\alpha(1-\alpha)\beta(1-\beta)\lt\dfrac{1}{16}\implies a\alpha(1-\alpha)a\beta(1-\beta)\lt\dfrac{a^2}{16}\\ \implies \dfrac{a^2}{16}\gt f(0)\cdot f(1)\implies |a|\gt4$$

$$\therefore a_{min}=5$$

Also, $$\because \qquad\qquad\alpha\cdot\beta\lt1\implies c\lt a_{min}\qquad\qquad\tag{4}$$

$$(2)\implies a+c\ge1+b\implies a_{min}+c-1\ge b\implies 4+c\ge b \\ \implies c^2+8c+16\ge b^2\implies c^2+8c+16 \gt 4a_{min}c\\ \implies c^2-12c+16\gt0\implies c\in(0,6-2\sqrt5)\cup(6+2\sqrt5,\infty)$$

But, as $c\lt a_{min}$, hence $c=1$

Now, from $$b^2\gt 4a_{min}c\implies b^2\gt20$$ Hence, $b_{min}=5$

So, the value of least positive integers $a$,$b$ and $c$ for which $f(x)$ has roots b/w $0$ and $1$ is $(a,b,c)=(5,5,1)$

Other answers are still welcome.

Solution 3:

Let $x_1, x_2$ be the two roots with $0\lt x_1\lt x_2\lt 1$. The graph of $f(x)=ax^2-bx+c$ is a parabola (with $a\gt 0$) whose vertex $V$ is given by $f\space'(x)=2ax-b=0$, i.e. $$V=\left(\dfrac{b}{2a},f\left(\dfrac{b}{2a}\right)\right)=\left(\dfrac{b}{2a},\dfrac{4ac-b^2}{4a}\right)$$ Because the roots are real we have $b^2-4ac\gt 0$, furthermore $f(x)$ is decreasing for $x\lt\dfrac{b}{2a}$ and increasing for $x\gt\dfrac{b}{2a}$ so one has that both $f(0)$ and $f(1)$ are positive (since $f(0)\le0\Rightarrow x_1\lt0$ and $f(1)\le 0\rightarrow x_2\gt1$) It follows we have the conditions $$\begin{cases}f(1)=0\iff a+c\gt b\\4ac\lt b^2\\f(0)=c\gt 0\end{cases}$$ we can use just $$\begin{cases}a+c\gt b\\4ac\lt b^2\end{cases}$$ from which we get directly $b\ge 3$. Taking the minimun value $c=1$ we find incompatibility for $b=3$ and $b=4$ but for $b=5$ we have $$\begin{cases}a+1\gt b\\4a\lt b^2\end{cases}\Rightarrow 4\lt a\lt\frac{25}{4}\Rightarrow 4\lt a\le 6$$ so have solutions $$(a,b,c)=(5,5,1),(6,5,1)$$ and it is easily verified that the equations $5x^2-5x+1=0$ and $6x^2-5x+1=0$ have roots according to the proposed problem.

Thus the required minimun values of $(a,b)$ is $\color{red}{(5,5)}$.

It is a simple question to verify that for values of $c\gt1$ there are not solution for $b\lt5$, say $b=4$, because we would deduce $a\gt 3-h$ and $a\lt \dfrac{4}{1+h}\le 2$