Too complex to be possibly real

Note that $p^{3}q^{3} = 25 + 2 = 27 \Longrightarrow (pq)^{3} = 27$ ,

which has $pq = 3$ as a real solution. (There are two complex solutions as well. These will be dealt with in footnote **).

Next, note that $p^{3} + q^{3} = 10$ , and that

$(p + q)^{3} = p^{3} + q^{3} + 3pq(p + q) = 10 + 9(p + q)$ .

Now by observation, $p + q = -2$ is a solution to this equation, and so

$(p + q)^{3} - 9(p + q) - 10 = ((p + q) + 2)((p + q)^{2} - 2(p + q) - 5) = 0.$

The other two roots are non-integral, (namely $1 \pm \sqrt{6}$ ), so the only possible integer value of $p + q$ is $\boxed{-2}$ .

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Note: Now, to be complete, we need to look at where the complex solutions for $pq$ lead us with regards to values for $p + q$ . We have the equation

$(pq)^{3} - 27 = 0 \Longrightarrow (pq - 3)((pq)^{2} + 3pq + 9) = 0$ ,

which besides $pq = 3$ has solutions $pq = -\dfrac{3}{2} \pm i*\dfrac{3\sqrt{3}}{2}.$

This results in the equations

$(p + q)^{3} + \dfrac{9}{2}(-1 \pm i*\sqrt{3})(p + q) - 10 = 0.$

Given that there are complex coefficients involved here, if $p + q$ were any real value (other than $0$ ) then there would be a non-zero complex component on the LHS of this equation but no complex component on the RHS. If $p + q$ were $0$ then the LHS would come out to $-10$ , thus $p + q \ne 0.$ Thus with $pq$ complex we would obtain only complex values for $p + q$ as a result, and thus $pq = 3$ is the only value that can, (and does), yield an integral value for $p + q.$

First, we find that $p^3 q^3 = 25+2= 27 = (pq)^3$ , which gives $pq=3$ , and that $p^3+q^3=10$ .

Now, the following is a simple identity:

$(p+q)^3 = p^3 + q^3 + 3pq(p+q)$ .

Substituting what we know yields:

$(p+q)^3=10+9(p+q)$ .

This is in fact a cubic equation of $p+q$ . To simplify notation, let's define $u=p+q$ and rewrite the equation:

$u^3-9u-10=0$ .

By the rational root theorem, we know that the only integer solutions possible are the factors of $10$ , which are $\pm 1, \pm 2, \pm 5, \pm 10$ .

A manual check shows that $u=-2$ is the only integer solution.

Thus, $\boxed{p+q=-2}$ .

Let $p = r_1 \text{ cis } \theta_1$ and $q = r_2 \text{ cis } \theta_2$ where $r_1, r_2 \in \mathbb{R}^+_0$ . Then $r_1^3 = |p^3| = \sqrt{5^2 + (\sqrt{2})^2} = \sqrt{27}$ and thus $r_1 = \sqrt{3}$ , and similarly with $q^3$ and $r_2$ .

Now note that $p+q = \sqrt{3} (\text{cis } \theta_1 + \text{cis } \theta_2)$ , and the only way it is an integer is when the imaginary part is zero; that is, $\sin \theta_1 + \sin \theta_2 = 0$ or $\theta_1 \equiv -\theta_2 \pmod{2\pi}$ . In other words, $p,q$ are conjugates of each other, and $p+q = 2 \text{ Re}(p)$ .

Finally, observe that $p = -1+\sqrt{2} i$ works by sheer luck. Thus the other two roots must be $(-1+\sqrt{2} i) \cdot \text{cis } \frac{2\pi}{3}$ and $(-1+\sqrt{2} i) \cdot \text{cis } \frac{4\pi}{3}$ , as $\frac{p}{-1+\sqrt{2} i}$ is a third root of unity. It can be easily shown that the other two solutions don't give a half-integer real part, thus $p = -1+\sqrt{2}i$ is the only one that works, and the answer is thus $p+q = 2 \text{ Re}(p) = \boxed{-2}$ .

We are given the following :-

$p^3=5+i\sqrt{2}...(i)$

$q^3=5-i\sqrt{2}...(ii)$

Multiplying $(i)$ and $(ii)$ :-

$p^3q^3=27$

Hence, $pq=3$

Adding $(i)$ and $(ii)$ :-

$p^3+q^3=10$

We know that $p^3+q^3=(p+q)(p^2-pq+q^2)=(p+q)((p+q)^2-3pq)$

Let $p+q=t$

Hence, $(t)(t^2-9)=10$

$t^3-9t-10=0$

From observation, $t$ comes out to be -2

Hence, the answer is $-2$

Solving for $p$ and $q$ , the easiest root to compute for both are the following:

$p_{0} = -1 + i\sqrt{2}$

$q_{0} = -1 - i\sqrt{2}$

The remaining roots can be found by multiplying the roots by the cubic roots of unity, thus giving us:

$p_{1} = \frac{1 -\sqrt{6}}{2} - i*\frac{\sqrt{3} + \sqrt{2}}{2}$

$p_{2} = \frac{1 + \sqrt{6}}{2} + i*\frac{\sqrt{3} - \sqrt{2}}{2}$

$q_{1} = \frac{1 + \sqrt{6}}{2} - i*\frac{\sqrt{3} - \sqrt{2}}{2}$

$q_{2} = \frac{1 -\sqrt{6}}{2} + i*\frac{\sqrt{3} + \sqrt{2}}{2}$

The only possible combination for which $p + q$ could be an integer was $p_{0} + q_{0} = -2$ .

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Now, just for clarification as to how I've found the results for $p_{0}$ and $q_{0}$ , I simply wrote $p = a + bi$ , cubed it and worked my way from there, assuming that $b = \sqrt{2}$ as a first guess and then I've found the other values. I'm well aware math is not a game of guessing, but if it works and you find the solution sought, it doesn't really matter, does it?