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Find the number of integers $k$ such that $$x^3-27x+k$$ has at least two distinct integer roots.



Here are some of the few thoughts that came to me on how to approach the problem:

  1. Using derivative to find the condition for existence of solutions. However, this leads to a very large bound on $k$
  2. Since the value of $k$ just shifts the graph of the function along the $y$ direction, we can find the bounds on the roots using the bounds on $k$ found above.

    The answer is there are $2$ integers $k$
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4 Answers 4

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Using the relations of the roots, we can get $$x^2+y^2+xy=27$$ where $x$ and $y$ are the 2 roots of the polynomial. Taking it as a quadratic in $x$, we get: $$x=\frac{-y \pm \sqrt{108-3y^2} }{2}$$ Since $x$ and $y$ are integers, we get that $108-3y^2$ is a perfect square which is satisfied only by $y \in \{\pm3,\pm6 \}$ which gives us all possible values of $k=\pm 54$

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Using elementary algebraic number theory it could go as follows.

Let the zeros of the cubic be $a,b,c$. By Vieta relations $c=-b-a$, so if $a,b$ are integers so is $c$. Comparing the linear terms of the factorization $$ (x-a)(x-b)(x-c)=x^3-27x+k $$ gives $$ a^2+ab+b^2=27.\qquad(*) $$ We recognize $a^2+ab+b^2$ as the norm $N(z)=z\overline{z}$ of the Eisensteinian integer $z=a-b\omega$ where $\omega=(-1+i\sqrt3)/2$ is the primitive third root of unity. The ring of Eisensteinian integers $R=\Bbb{Z}[\omega]$ has unique factorization. Furthermore, we know that the rational prime $3$ ramifies in $R$, so up to a unit factor there is only a single element of $R$ with norm $3^3=27$. As $a=b=3$ is a solution of $(*)$ and $-\omega$ generates the group of units of $R$ this implies that all the solutions of $(*)$ are $$ z=a-b\omega=\pm(3-3\omega)\omega^j, $$ for some $j=0,1,2$. The possibilities are thus $(a,b)=\pm(3,3)$, $(a,b)=\pm(-3,6)$ and $(a,b)=\pm(6,-3)$.

So $(a,b,c)=\pm(3,3,-6)$ up to a permutation. By Vieta relations $$ k=-abc=\pm 54. $$

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Let $f(x)=x^3-27x+k$. By considering the derivative of $f$, we see that $f(x)$ attains a local maximum of $f(-3)=54+k$ at $x=-3$, and a local minimum of $f(3)=-54+k$ at $x=3$. Furthermore, $f$ is strictly increasing on the intervals $(-\infty,-3]$ and $[3,\infty)$, and strictly decreasing on the interval $[-3,3]$.

If $k<-54$, then $f(3)<f(-3)<0$, so $f$ is negative on $(-\infty,3]$ and has exactly one zero on $[3,\infty)$ (since $f$ is strictly increasing on this interval). Thus, if $k<-54$, then $f$ has only one zero. Similarly, if $k>54$, then $f(-3)>f(3)>0$, and so $f$ has only one zero, which is attained on the interval $(-\infty,3]$.

Therefore, we must have that $-54\leqslant k\leqslant54$. Observe that we now have that $f(-3)\geqslant0\geqslant f(3)$. Thus $f$ must have exactly one zero in the interval $[-3,3]$ (since $f$ is strictly decreasing on this interval). Note that $f(0)=0$ if and only if $k=0$, in which case the only integer root of $f$ is $0$ (since $27$ is not an integer square). From now on assume $k\neq0$.

Suppose $f$ has at least $2$ integer roots and let $a,b,c$ be the (not necessarily distinct) $3$ real roots of $f$ (no non-real roots since they come in pairs). We have that $-k=abc$. Suppose $a,b$ are integers. Since $k\neq0$, we see that $a,b\neq0$, so $c=-k/ab$ is rational. As $f$ is a monic polynomial with integer coefficients, all of its rational roots are integers. Therefore, $f$ has $3$ (not necessarily distinct) integer roots.

Recall that $f$ has exactly one distinct zero, say $a$, which lies in $[-3,3]\setminus\{0\}$. Furthermore, this zero $a$ has multiplicity 1 (i.e. is the unique zero of $f$ in $[-3,3]\setminus\{0\}$) if $f'(a)\neq0$, i.e if $|a|\neq3$. If $|a|=3$, $f(x)=(x\pm 3)^{2}(x\mp6)$. This shows that $k=54$ and $k=-54$ are valid solutions.

If $|a|=1$, $f(x)=(x\pm 1)(x^{2}\mp x- 26)$. If $|a|=2$, $f(x)=(x\pm 2)(x^{2}\mp 2x- 23)$. In these two cases we must have $|bc|\in\{23,26\}$, where $b,c$ are the remaining roots of $f$ and must therefore satisfy $|b|,|c|>3$ (since $a$ is a root of multiplicity 1). By factorising $23$ and $26$ into primes, we see no such integers $b,c$ can exist.

Therefore, we must have $k=54$ or $k=-54$.

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For $x^3+px+q=0$ we need $\frac{p^3}{27}+\frac{q^4}{4}\leq0,$ which gives $-54\leq k\leq 54$ and checking.

There is a bit of better reasoning.

Let $a$ be one of roots.

Thus, $x^3-27x=a^3-27a$ or $(x-a)(x^2+ax+a^2-27)=0$, which says $a^2-4(a^2-27)\geq0$, which gives $-6\leq a\leq 6$ and checking again.

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  • $\begingroup$ You need to find the values of k such that there exist at least 2 distinct integer roots $\endgroup$ Jul 28, 2017 at 16:04
  • $\begingroup$ I was honestly thinking, just use the fact that 1 and 27 are odd, and show that if k is odd we end up with an odd number when x is odd. if k is odd when x is even we also end up with an odd number. 0 is not odd. therefore all k values that give solutions must be even. $\endgroup$
    – user451844
    Jul 28, 2017 at 16:05
  • $\begingroup$ @Transformator Yes of course! For $k=54$ we have two distinct roots. $3$ and $-6$. $\endgroup$ Jul 28, 2017 at 16:06
  • $\begingroup$ @MichaelRozenberg But know how can you show that every other value of $k$ in said range does not satisfy the conditions? $\endgroup$ Jul 28, 2017 at 16:07
  • $\begingroup$ @Transformator It's just checking, $\endgroup$ Jul 28, 2017 at 16:10

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