Must be fixed q or r, because (p,q,r) and (pc,qr,cr) both satisfyed your property.

I don't think there's a problem. Note that what we want is the minimal integral value of $p$ (which must exist).

http://www.mathlinks.ro/Forum/viewtopic.php?t=24806

Let $0<a<b<1, q=p(a+b),r=pab,$ p any positive satisfyed your condition. Has not minimum. inf(p)=0.

To Rust: $p$, $q$, $r$ must be positive integers!!

Possibly a more rigorous solution? Let the quadratic be $ax^2-bx+c$ where $a,b,c$ are positive integers. It suffices to find $a,b,c$ such that $\frac{b-\sqrt{b^2-4ac}}{2a} > 0$ $\frac{b+\sqrt{b^2-4ac}}{2a} < 1$ and $a$ is as small as possible. The first equation always holds so rewrite the second as $b+ \sqrt{b^2-4ac} < 2a$ We note that $b^2 > 4ac \geq 4a$ since $c \neq 0$ (roots are distinct) so ... For $a = 1$ note $b \geq 3$ so the LHS is at least $3+\sqrt{3^2-4} > 2$ For $a = 2$ note $b \geq 3$ so the LHS is at least $3+\sqrt{3^2-8} = 4$ For $a = 3$ note $b \geq 4$ so the LHS is at least $4+\sqrt{4^2-12} = 6$ For $a = 4$ note $b \geq 5$ so the LHS is at least $5+\sqrt{5^2-16} = 8$ For $a = 5$ we may take $b = 5$ and $c = 1$ and verify that $5 + \sqrt{5^2-20} < 10$. So minimal $a = 5$ as shown by the quadratic $5x^2 - 5x + 1 = 0$.

ThAzN1 wrote: It suffices to find $a,b,c$ such that $\frac{b-\sqrt{b^2-4ac}}{2a} > 0$ $\frac{b+\sqrt{b^2-4ac}}{2a} < 1$ and $a$ is as small as possible. You forgot the end braces for \frac, by the way.

letting $f(x)=px^2-qx+r$ then $f(1)>0$ so $p+r>q$ , $q^2>4pr$ also the minimum point is $\frac{q}{2p}<1$ then we can find $p>r$ so $2p>p+r>q$ and we get $2p\geq q+2$ also $p>r\geq 1$ so $p\geq 2$ when $p=2$ , $q^2>8$ and so $q\geq 3$ ( does satisfy $q<2p$) when $p=3$ , $q^2>12$ and $q\geq 4$ (doesnt satify $p+r>q$) when $p=4$ , $q\geq 5$ (doesnt satisfy $p+r>q$) when $p=5$ , $q\geq 5$ and this one satisfy the condition . So the min for $p=5$