I recently happened upon a problem that I unfortunately am not able to make any progress on. Here it is:
Let be integers.
If , prove that must be perfect squares.
I also asked this on yahoo answers hereProve that p, q and r must be perfect squares? - Yahoo! Answers .But, I don't think that those solutions address the possibility of having three or two irrational numbers that add up to an integer.
Thank you. Any help with this problem is appreciated.
This is a simple case of the following argument:
Let such that . Then
Proof:
We'll prove that . By symmetry it will then apply to and .
Let . Then
Then from my post here we can deduce that . For convenience I'll prove it here:
Therefore
and we're done (under the assumption that . Otherwise and the result is trivial)
I may have overlooked something, since I am not really familiar with set theory at all, but I do not know the statement that you have just proved. Sorry, it's because I'm quite a few years younger than university age. I can understand the proof, but I do not know what you have proved.
What I've proved is the statement before the word "Proof":
if are rational numbers and is also rational,
then all those square roots are rational themselves.
Applied to your problem - we're given that are integers
and is also an integer. Since being
an integer is just a particular case of being rational, the above argument shows
that all are rational. From here,
can you deduce that are integers?
(note that a square root of an integer is integer or irrational. If you didn't know this fact,
you may wanna give it a try and prove it)
It came from noticing that if is rational then so is (when are rational).
But then, since we know that sum of rational numbers is rational we conclude that is rational.
Dividing by 2 (since it's rational) yields . The same logic was applied
in the above proof, but for and - we see that it gives us
and we can divide out the . What I did was just
to squeeze all these steps together.
Two short ways I know:
- in the same fashion the square root of 2 is proved to be irrational
- by the Rational Root Theorem applied to the polynomial . From this, if is rational then it's integer (assuming is integer), which is what we want.