# Integral roots of cubic equation $x^3-27x+k=0$

The number of integers $$k$$ for which the equation $$x^3-27x+k=0$$ has atleast two distinct integer roots is

(A)$$1$$ (B)$$2$$ (C)$$3$$ (D)$$4$$

My Attempt: The condition for cubic $$x^3+ax+b=0$$to have $$3$$ real roots happens to be $$4a^3+27b^2\leq0$$. But how to go about finding condition for integer roots.

• math.stackexchange.com/questions/2157643/… Commented Apr 1, 2019 at 3:41
• I did apply the Cardano's method. The only way to do is i suppose by taking $x=-3,-2,-1,0,1,2,3$ and check whether we get another integer root and at that root k should be an integer. Commented Apr 1, 2019 at 3:51
• May be you can use rational root theorem
– ersh
Commented Apr 1, 2019 at 4:45
• The condition that k is an integer does not worth mentioning it follows from vieta relations. Commented Apr 1, 2019 at 5:36

Suppose $$x^3 - 27x + k = 0$$ has distinct integer roots $$a$$ and $$b$$; then $$a^3 - 27a = b^3 - 27b,$$ or $$a^3 - b^3 = 27(a - b).$$ Since, by hypothesis, $$a\ne b$$, a factor of $$a-b$$ can be removed, resulting in $$a^2 + ab + b^2 = 27.$$ After multiplying by $$4$$, this can be rearranged into $$(2a + b)^2 + 3b^2 = 108.$$ It follows that the integer $$2a+b$$ is a multiple of $$3$$, and has a square $$\le 108$$; thus $$2a+b = 0,\pm3,\pm6$$ or $$\pm9$$.

• If $$2a+b = 0$$, then $$b^2 = 36$$, so $$b = \pm6$$.
• If $$2a+b = \pm3$$, then $$b^2 = 33$$, so this has no integral solution.
• If $$2a+b = \pm6$$, then $$b^2 = 24$$, so this has no integral solution.
• If $$2a+b = \pm9$$, then $$b^2 = 9$$, so $$b = \pm3$$.

In the first case, we find $$(a,b) = (-3,6)$$ or $$(3,-6)$$. In the fourth case, the four possible combinations of signs result in $$(a,b) = (3,3), (6,-3), (-3,-3)$$ or $$(-6,3)$$. Rejecting the cases with $$a=b$$, $$(a,b) = (-3,6)$$ or $$(6,-3)$$ results in $$k = 27a - a^3 = -54$$ and $$(a,b) = (3,-6)$$ or $$(-6,3)$$ results in $$k = 54$$. Thus there are two possible values of $$k$$.

If $$x=b$$ is one of the solutions

$$k=27b-b^3$$

Now if $$b$$ is a repeated root and $$c$$ is the third one,

$$0=b+b+c\iff c=-2b$$

$$\implies x^3-27x+k=(x-b)^2(x+2b)=x^3+x(2b+b^2)-2b^3$$

$$\implies b^2+2b=-27\iff b^2+2b+27=0$$ which does not have an integer solution.

So, we can not have repeated roots.

The rest two solutions will be available from the quadratic equation $$0=\dfrac{x^3-27x-(b^3-27b)}{x-b}=x^2+bx+b^2-27$$

As $$x$$ is an integer, the he discriminant must be perfect square i.e.,

$$b^2-4(b^2-27)=108-3b^2=3(36-b^2)=D$$(say)

$$\implies36-b^2\ge0\iff b^2\le36\implies b\le6$$

Also $$3$$ must divide $$b$$ to keep $$D$$ perfect square

So, $$b\in[0,\pm3,\pm6]$$

Clearly, $$D$$ is perfect square only for $$b=\pm6$$ .

Here is another solution mainly from vieta relations:-

Say the roots be $$\large p,q,r$$ all are integers since there sum is zero two integer implies the third one to be so

Now $$pqr=k \\ pq+qr+rp=-27 \\ p+q+r=0$$ $$\implies pq(p+q)=k \\ and \\ pq-(p+q)^2=-27 \\ \implies p^2+q^2+pq=27 \\ p=\frac{-q+\sqrt{3(36-q^2)}}{2}\\or,\\p=\frac{-q-\sqrt{3(36-q^2)}}{2}$$ arriving at this the number of integral solution is not very difficult to count first of all we conclude $$\large q$$ is even and others such conclusion reduces the cases and the solution follows.