MHB Proving Rationality of Squares of 9th Day 2 ARO 2004/2005 Numbers

[Smb]

Ten mutually distinct non-zero reals are given such that for any two, either their sum or their product is rational. Prove that squares of all these numbers are rational.
I tried using 3 of those numbers - a, b and c. And I checked each of the possible situations but I'm not sure if my maths teacher is going to accept it.
The problem is from the All-Russian Olympiad 2004/2005 9th Day 2.

 

[Joppy]

Is this a homework assignment? If so, I recommend you post as much working as possible.

 

[Smb]

Is this a homework assignment? If so, I recommend you post as much working as possible.

I worked out the answer easily but the method is slow.
I'll write for the first one because I don't have much time now:
a+b(-Q
b+c(-Q
a+c(-Q
ab+bc+ac+b^2(-Q
ab+bc+ac+c^2(-Q
(b-c)(b+c)(-Q
b-c(- Q Analogically we can do for each pair of numbers.
b^2-2bc+c^2-B^2-2bc-c^2(-Q
=> bc (- Q Analogically we can do for each pair of numbers. =>
b^2(- Q Analogically we can do for each pair of numbers.

 

[I like Serena]

I worked out the answer easily but the method is slow.
I'll write for the first one because I don't have much time now:
a+b(-Q
b+c(-Q
a+c(-Q
ab+bc+ac+b^2(-Q
ab+bc+ac+c^2(-Q
(b-c)(b+c)(-Q
b-c(- Q Analogically we can do for each pair of numbers.
b^2-2bc+c^2-B^2-2bc-c^2(-Q
=> bc (- Q Analogically we can do for each pair of numbers. =>
b^2(- Q Analogically we can do for each pair of numbers.

Hi Smb! Welcome to MHB! (Smile)Whatever way I can think of, it seems to be a number of cases, although I think we can do with fewer than you are suggesting.
Still, if we could cover all 8 similar cases we would be done, although I don't think we can.For starters, what you have now cannot occur, since if we start with $\def\inQ{\in\mathbb Q}
a+b, a+c \inQ$, then it follows that $(a+b)-(a+c) = (b-c) \inQ$.
Now suppose $b+c\inQ$, then $\frac 12((b+c)+(b-c))=b\inQ$ and similarly $c \inQ$, so that $bc\inQ$, which contradicts the either-or.
Therefore $b+c\not\inQ$.Let's start with assuming that we can find a triplet $(a,b,c)$ with $ab,ac\inQ$.
Then $\frac {ab}{ac}=\frac bc \inQ$, isn't it?
Now suppose $b+c\inQ$, then $\frac bc\cdot c + c = (\frac bc + 1)c \inQ$.
Can we figure out from there if any of $b$, $c$, and in particular $bc$ is rational? (Wondering)
And if so, can we find that $a^2\inQ$?

 

[Smb]

Hi Smb! Welcome to MHB! (Smile)Whatever way I can think of, it seems to be a number of cases, although I think we can do with fewer than you are suggesting.
Still, if we could cover all 8 similar cases we would be done, although I don't think we can.For starters, what you have now cannot occur, since if we start with $\def\inQ{\in\mathbb Q}
a+b, a+c \inQ$, then it follows that $(a+b)-(a+c) = (b-c) \inQ$.
Now suppose $b+c\inQ$, then $\frac 12((b+c)+(b-c))=b\inQ$ and similarly $c \inQ$, so that $bc\inQ$, which contradicts the either-or.
Therefore $b+c\not\inQ$.Let's start with assuming that we can find a triplet $(a,b,c)$ with $ab,ac\inQ$.
Then $\frac {ab}{ac}=\frac bc \inQ$, isn't it?
Now suppose $b+c\inQ$, then $\frac bc\cdot c + c = (\frac bc + 1)c \inQ$.
Can we figure out from there if any of $b$, $c$, and in particular $bc$ is rational? (Wondering)
And if so, can we find that $a^2\inQ$?

I forgot to rewrite it! I'm sorry for the misunderstanding it can be and/or.Both b +c(-Q and bc(-Q could be a case..

 

[I like Serena]

I forgot to rewrite it! I'm sorry for the misunderstanding it can be and/or.Both b +c(-Q and bc(-Q could be a case..

Fair enough. The proof doesn't really change though. Just a couple more cases to consider.