Inequality involving a quadratic equation

Question

Let $a,b,c$ be integers and suppose the equation $$f(x) = ax^2 + bx + c = 0$$ has an irrational root $r$ . Let $u=\dfrac{p}{q}$ be any rational no. such that $|u-r|<1$.

Prove that $$\dfrac{1}{q^2} ≤ |f(u)|≤ K|u-r|$$ for some constant $K$ . Deduce that there is a constant $M$ such that $\bigl|r\dfrac{p}{q}\bigr| ≥ M/q^2$.

Obviously, this gives an upper bound. For the lower bound, ponder on what kind of number $ap^2+bpq+cq^2$ must be. And can you check if you missed a minus sign in your last statement? — Lutz Lehmann, May 28 '14 at 11:21
Then there is an earlier error in the reproduction chain of the task. It should be $|r-\tfrac pq|\ge M/q^2$. See also Liouville theorem (one of the many). — Lutz Lehmann, May 28 '14 at 11:26
And the better, more strict, assumption would be that the rational number is the best approximation with denominator $q$, i.e., $|qr-p|<1$. — Lutz Lehmann, May 28 '14 at 11:36

Lutz Lehmann · Accepted Answer · 2014-05-28T12:11:20.780

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This task is to prove one elementary statement about Diophantine approximation.

If $f(x)$ is an univariate polynomial of degree $d$ with integer coefficients, $r$ an irrational root and $u=\frac pq$, $p,q\in \Bbb Z$, $q>0$, with $|r-u|<1$ not itself a root, then $$ q^df(u) $$ is a non-zero integer, and thus $$ \frac1{q^d}\le|f(u)|=|f(u)-f(r)|=|u-r|\,|f'(r+\theta(u-r))| $$ Let $K$ be a bound on the first derivative, $|f'(x)|\le K$ on the interval $x\in [r-1,r+1]$, then $$ |u-r|\ge \frac1{K\,q^d} $$ is a result by Liouville around 1840.

edited May 28 '14 at 12:11

answered May 28 '14 at 11:57

Lutz Lehmann

126,666

Winner of the national mathematical olympics. And $n=11$ is the only solution. – Lutz Lehmann May 28 '14 at 12:43

Inequality involving a quadratic equation

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