# A “complex” complex number problem

$a,b,c$ are cube roots of $p$ ,($p<0$) then for any permissible value of $x,y,z$ which is given by

$$\frac{|xa+yb+zc|}{|xb+yc+za|} + (a_1^2-2b_1^2)\omega + \omega^2([x]+[y]+[z]) = 0$$

$\omega$ is cube root of unity $a_1,b_1$ are real positive numbers and $b_1$ is prime . We have to find the value of $[x+a_1] + [y+b_1] + [z]$ . where $[\cdot ]$ denotes the greatest integer function .

I don't have any idea , where to start !

• You mean, $x, y, z$ is permissible is defined by saying "the relation holds"? – Igor Rivin Dec 3 '13 at 17:30
• So, this result is independent of $p$? – Doc Dec 3 '13 at 17:40
• I think the result is just a number . – abkds Dec 3 '13 at 17:53
• Yes, the result is just a number. But note, that number is independent of $p$, $a$, $b$, and $c$. Okay, I guess it must depend on these values implicitly, so never mind. – Doc Dec 3 '13 at 18:23
• Let $N$ be the number you want. I get $N=b_1(b_1+1)+\frac{a_1-1}{2}$. However, (1) I assumed that $a_1$ was an integer; (2) I didn't use the fact that $b_1$ is a prime; (3) I could well have messed up as the calculation was painful. Note also, that my "result" implies that $a_1$ is odd (which looks a bit strange, but may actually have to be the case, given the equality restriction). – Doc Dec 3 '13 at 19:13

We know that one of the following holds:

• $b = a\omega$ and $c = b\omega$ and $a = c\omega$.
• $b = c\omega$ and $a = b\omega$ and $c = a\omega$.

In each of these cases, the ratio

$$\frac{xa+yb+zc}{xb+yc+za}$$

has a definite value independent of $p$, namely $\omega^2$ or $\omega$ respectively. In both cases,

$$\frac{|xa+yb+zc|}{|xb+yc+za|} = \left|\frac{xa+yb+zc}{xb+yc+za}\right| = 1$$

Now we have an equation of the form $p + q\omega + r\omega^2 = 0$. From this we may deduce $p = q = r$. In the case in point:

$$a_1^2 - 2b_1^2 = 1$$ $$[x]+[y]+[z] = 1$$

At this point I can make no progress without assuming that $a_1$ is an integer. I will assume that that's what the question intended.

Then $a_1$ is odd, say $a_1 = 2k+1$, so

$$4k^2 + 4k + 1 - 2b_1^2 = 1$$ $$2k(k+1) = b_1^2$$

Therefore $b_1$ is even, and since it is prime, it must be $2$, and $a_1$ must therefore be $3$. Then

$$[x+a_1] + [y+b_1] + [z]$$ $$= [x] + 2 + [y] + 3 + [z]$$ $$= ([x] + [y] + [z]) + 5$$ $$= 6$$

• Indeed that is the answer . Very well done :) . Thanks – abkds Dec 6 '13 at 5:00
• Where did you find this question? It is clearly a deliberately obfuscated puzzle! – apt1002 Dec 6 '13 at 5:09
• Its a question from a book ( for preparing for competitive exam ) . I was just solving those problems – abkds Dec 6 '13 at 5:12