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 $i$?

  • $\begingroup$ Oh i'm sorry i miss the fact that $a_i$'s are all non-zeros. $\endgroup$ – D. N. Nov 19 '13 at 4:27
  • $\begingroup$ Now i will edit the questions. $\endgroup$ – D. N. Nov 19 '13 at 4:28
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    $\begingroup$ If you draw a graph of $K_5$ with the $a$'s, the additions and multiplications must form the Petersen graph, because if you have a triangle with the same sign you can use that to prove all three elements of the triangle rational, then you can get the other two. $\endgroup$ – Ross Millikan Nov 19 '13 at 4:28
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    $\begingroup$ I got on this thinking pigeonhole principle. If you have $a+b, a+c, b+c$ rational, then $a-b, a+b$ are rational, so $a,b$ rational, then $a+d$ or $ad$, so $d,e$ rational. The Petersen graph is a good way to represent the intermediate stage, but you don't have to reference it. $\endgroup$ – Ross Millikan Nov 19 '13 at 4:39
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    $\begingroup$ @RossMillikan: and the conclusion is...? It's not quite clear to me how you get the Petersen graph, or what you do with it. $\endgroup$ – Robert Israel Nov 19 '13 at 5:46

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 $a_5$.

  • $\begingroup$ Oops: it's not so obvious that a blue $3$-cycle makes the other two $a_i^2$ rational. I'm pretty sure I can fix the gap, but I don't have time right now. $\endgroup$ – Robert Israel Nov 19 '13 at 16:37
  • $\begingroup$ @RobertIsreal I Just dont understand, that, why it is not possible to colour the edges of $K_5$ in two colours so there are no odd cycles of either colour? Thanks for your time. $\endgroup$ – D. N. Dec 9 '13 at 5:58
  • $\begingroup$ Try it and see! $\endgroup$ – Robert Israel Dec 9 '13 at 15:51
  • $\begingroup$ ... or enumerate the cases. Only 10 edges, and wlog you can assume 1-2 and 2-3 are red and 1-3 is blue. $\endgroup$ – Robert Israel Dec 9 '13 at 15:54
  • $\begingroup$ Ok i wil try. By the way, thank you. $\endgroup$ – D. N. Dec 9 '13 at 16:27

No. Take $a_1=\pi, a_2=\frac{1}{\pi}, a_3=-\pi, a_4=\frac{-1}{\pi}, a_5=0$.

Then $a_1a_2 = 1, a_1+a_3=0,a_1a_4=-1,a_1a_5=0,a_2a_3=-1,a_2+a_4=0,a_2a_5=0,a_3a_4=1,a_3a_5=0, \text{and } a_4a_5=0.$

  • $\begingroup$ Good answer to the original question. $0$ is special. $\endgroup$ – Ross Millikan Nov 19 '13 at 4:29

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