Existence of five real numbers satisfying a given condition. Let $a_1,\dots,a_5$ be five distinct non-zero real numbers. Suppose that for $i\neq j$ either $a_i+a_j$ or $a_ia_j$ or both are rational numbers, does it implies that $a_i^2$ are rational numbers for all $i$?
 A: Consider
the complete graph $K_5$ on $5$ vertices.  Colour the edge $(i,j)$ blue if $a_i + a_j$ is rational, otherwise red.  
Suppose there is a red $m$-cycle $(i_1, i_2, \ldots, i_m$, i.e. 
$r_1 = a_{i_1} + a_{i_2}, r_2 = a_{i_2} + a_{i_3}, \ldots, r_m = a_{i_m} + a_{i_1}$ are all rational.  Then $a_{i_2} = r_1 - a_{i_1}$, $a_{i_3} = r_2 - r_1 + a_{i_1}$, etc, determining each $a_{i_j}$ in terms of the $r_j$ and $a_{i_1}$.  If $m$ is odd, when we come around the whole cycle we
get $a_{i_1} = r_{i_m} - r_{i_{m-1}} + \ldots + r_{i_1} - a_{i_1}$ 
which makes $a_{i_1}$  rational, and then 
we find that all $a_i$ are rational.
Similarly, if there is a blue $m$-cycle with $m$ odd, we would get
$a_{i_1} = r_{i_m} r_{i_{m-1}}^{-1} \ldots r_{i_1} a_{i_1}^{-1}$, which
makes $a_{i_1}^2$ rational, and then all $a_{i}^2$ are rational.
So in order to have an example with $a_i^2$ not all rational, we have to
be able to colour the edges of $K_5$ in two colours so there are no odd
cycles of either colour.  But it seems this is impossible.  So the answer is
yes, all $a_i^2$ must be rational.
EDIT: Here is the fix for the "gap":
If, say, $a_1 a_2, a_3$ is a blue triangle,  $a_1^2$, $a_2^2$ and $a_3^2$ are rational.  Now consider $a_4$.  If $a_1 a_4$  is rational, 
$a_4^2 = (a_1 a_4)^2/a_1^2$ is rational.  Similarly if $a_2 a_4$ is rational.  So suppose $a_1 + a_4 = r_{14}$ and $a_2 + a_4 = r_{24}$ are rational.  Then $a_1 - a_2 = r_{14} - r_{24}$  is rational (and nonzero, because the $a_i$ are distinct).  But then 
$a_1 + a_2 = \dfrac{a_1^2 - a_2^2}{a_1 - a_2}$ is rational, so $a_1$ and $a_2$ are rational, and $a_4$ is rational.
Similarly for $a_5$. 
A: No. Take $a_1=\pi, a_2=\frac{1}{\pi}, a_3=-\pi, a_4=\frac{-1}{\pi}, a_5=0$.
Then $a_1a_2 = 1, a_1+a_3=0,a_1a_4=-1,a_1a_5=0,a_2a_3=-1,a_2+a_4=0,a_2a_5=0,a_3a_4=1,a_3a_5=0, \text{and  } a_4a_5=0.$
