A quadratic trinomial $p(x)$ with real coefficients is given. Prove that there is a positive integer $n$ such that the equation $p(x) = \frac{1}{n}$ has no rational roots.
Problem
Source: International Zhautykov Olympiad 2013 - D2 - P1
Tags: quadratics, number theory, relatively prime, algebra unsolved, algebra
mlequi
17.01.2013 17:50
Let $p(x)=ax^2+bx+c$ and we assume the contrary that for all $n$, we have $p(x)=\frac{1}{n}$ has a rational root $u_n$.
Note that $au_k^2 + bu_k + c = \frac{1}{k}$ and $au_k^2 + bu_k + c = \frac{1}{l}$, so we have $a(u_k^2 - u_l^2) + b(u_k-u_l) = \frac{1}{k} - \frac{1}{l}$. Clearly, $u_k \neq u_l$ (for the contrary would result in $k=l$) so factorising, we have
$a(u_k+u_l)+b = \frac{\frac{1}{k}-\frac{1}{l}}{u_k-u_l}$. The point is,
\[ a(u_k+u_l) + b \]
is always rational for different $k$ and $l$.
Now, we have $a(u_1+u_2)+b$ and $a(u_1+u_3)+b$ are rationals, so their difference is also a rational number. Therefore, $a(u_2-u_3)$ is a rational number. Since $u_2 \neq u_3$, we have $a$ is a rational number. It follows that $b$ is a rational number since $a(u_1+u_2)+b$ is rational. Thus, since $ax^2+bx+c=1$ has a rational solution, $c$ must be rational.
Now, consider $ax^2+bx+c = \frac{1}{p}$ with $p$ be a prime that does not divide either numerator nor denominator of $a$ and $b$ when written in its lowest form. Needless to say, there are infinitely many of such primes.
Let $c = \frac{u}{v}$, $\gcd(u,v)=1$. For such prime $p$, suppose $x=y \frac{1}{p}$ is a solution, where $p$ does not divide the numerator of $y$ when written in lowest form. Then we have $\frac{ay^2}{p^2} + b\frac{y}{p} + c-\frac{1}{p}=0$ so we have $p|v$. There are other ways, suppose $x$'s denominator has no factor of $p$ when written in lowest form. Then we must have $apx^2+bpx+pc=1$. Since the chosen $x$'s denominator is not divisible by $p$, so the numerator of $apx^2+bxp$ is divisible by $p$. It turns out that $p$ divides the numerator of $pc-1$, in other words, $p|pu-v$, i.e $p|v$.
In both cases, we have $p|v$. Since there are infinitely many of such $p$'s, we must have $v=0$. But this is impossible, since $v$ is the denominator of $c$. Thus, our assumption is wrong and the statement of the problem is correct.
dinoboy
19.01.2013 03:40
Easier Suppose it were false, take $p(x)$ as a counterexample. By Lagrange Interpolating, $p(x)$ has rational coefficients. Thus let $p(x) = \frac{a}{b}x^2 + \frac{c}{d}x + \frac{e}{f}$ for $a,b,c,d,e,f$ integers. Take a prime $q > \max(a,b,c,d,e,f)$. Let $p(k) = \frac{1}{q}$. It is easy to show $q$ divides the denominator of $k$. But remark that then $v_q \left (\frac{ak^2}{b} + \frac{ck}{d} + \frac{e}{f} \right ) = 2v_q(k) \neq -1$ for all $k$ so we are done.
liimr
19.01.2013 06:30
that is easy to prove that coefficients should be rational so after denoting a by x/y , b by z/w and c by t/s as for rational coefficients, if one root is rational other should also be so which means discriminant should be perfect rational square and take n as y•w•s•k where k is any natural number then we have k^2•(N)+k•(M)=A^2 where N,M,A being natural numbers but k(N•k+M)=A^2 is not every time true (take any which is not perfect square and relatively prime to M) we have contradiction
Algie
19.01.2013 18:29
Suppose $af(n)^{2}+bf(n)+c- \frac{1}{n}=0$ for each positive integer $n$. ($f(n)$ is rational) $a(f(n)-f(m_{1}))(f(n)+f(m_{1}))+b(f(n)-f(m_{1})) - \frac{1}{n} + \frac{1}{m_{1}} = 0$, so $a(f(n)+f(m_{1}))+b$ is rational for $n \neq m_{1}$. $a(f(n)+f(m_{2}))+b$ is rational, so $a(f(m_{1})-f(m_{2}))$ is rational, so $a$ is rational. Then we easy get that $b$ and $c$ are also rational. Note that there exists such $n_{0}$ that $an_{0}=k>1$, $bn_{0}=m$, $cn_{0}=l$ are integers and $k$ is not a perfect square. Then for each positive integer $t$ there exists such rational $x$ that $ax^{2}+bx+c- \frac{1}{n_{0}t}=0=tkx^{2}+tmx+lt-1$ Thus $D = t^{2}(m^2-4kl)+4kt$ is a perfect square for each positive integer $t$. Let $t=(4k)^2$. Then $D=(4k)^{3}(4km^{2}-(4k)^{2}l+1)$, which is not a perfect square, because $ \nu_{k}(D)=3$. Contradiction.
anantmudgal09
09.09.2016 21:11
Claim No such $P$ (non constant) exists. Proof Assume not. Then, by the Lagrange Interpolation formula, we get that $P$ has rational coefficients. Thus, we may write $P(X)=\frac{f(X)}{N}$ for some integer $N>0$ so that $f$ has integer coefficients. Now, we see that there exists a sequence of rational numbers $(a_n)_{n \ge 1}$ such that $f(a_n)=\frac{N}{n}$ for all $n$. Let $f(X)=aX^2+bX+c$ where $a \not= 0$ is an integer. Now, write $a_n=\frac{p_n}{q_n}$ for coprime integers $p_n,q_n$ and non zero denominator. Choose $n=r>\max{N,|a|}$ to be a prime number. We have \begin{align*} r(ap_n^2+bp_nq_n+cq_n^2)=Nq_n^2 \end{align*}and thus, $r \mid q_n$. Therefore, $r \mid ap_n^2$ and so $r \mid a$, a contradiction. The conclusion holds. Comment This result holds for all non constant polynomials $P$ and the same argument may be applied.
lazizbek42
13.12.2021 13:41
$p(X)=aX^2+bX+c$ first we prove that a, b, c are rational. Suppose that for an arbitrary n let be a rational root in this equation. $ap^2+bp+c-\frac{1}{n}=0$ $aq^2+bq+c-\frac{1}{m}=0$ $a(p^2-q^2)+b(p-q)+\frac{1}{m}-\frac{1}{n}=0$ $a(p+q)+b=\frac{m-n}{mn(p-q)}$ if we do this again and divide the two equations, it turns out that a is rational.so b is also rational.so c is also rational. in general $p(x)=\frac{ex^2+fx+k}{N} we can say that. The following equation for an arbitrary n natural has a rational root.Hence D is also a definite square. $nex^2+nfx+nk-N=0$ We choose a prime number large enough for n. D will not be a clear square.