How do you prove that p, q and r must be perfect squares?

**JRichardson1729** · October 17th 2010, 08:47 AM

I recently happened upon a problem that I unfortunately am not able to make any progress on. Here it is:

Let $p, q, r, s$ be integers.

If $\sqrt{p}+\sqrt{q}+\sqrt{r}=s$ , prove that $p, q, r$ 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.

**wonderboy1953** · October 17th 2010, 09:27 AM

Originally Posted by JRichardson1729

I recently happened upon a problem that I unfortunately am not able to make any progress on. Here it is:

Let p, q, r, s be integers.

If $\sqrt{p}+\sqrt{q}+\sqrt{r}=s$ , prove that p, q, r 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.

Unless something is missing from this problem, it's not true in general (many counterexamples - please recheck).

**JRichardson1729** · October 17th 2010, 10:39 AM

Originally Posted by wonderboy1953

Unless something is missing from this problem, it's not true in general (many counterexamples - please recheck).

I am reasonably sure that it is right. Could you please give a counterexample?

**Unbeatable0** · October 18th 2010, 07:11 AM

This is a simple case of the following argument:

Let $a,b,c \in \mathbb{Q}$ such that $\sqrt{a}+\sqrt{b}+\sqrt{c}\in\mathbb{Q}$ . Then $\sqrt{a}, \sqrt{b}, \sqrt{c} \in \mathbb{Q}$

Proof:

We'll prove that $\sqrt{a}\in\mathbb{Q}$ . By symmetry it will then apply to $\sqrt{b}$ and $\sqrt{c}$ .

Let $k = \sqrt{a}+\sqrt{b}+\sqrt{c}$ . Then

$(k-\sqrt{a})^2 = (\sqrt{b}+\sqrt{c})^2$

$k^2-2k\sqrt{a}+a = b+c+2\sqrt{bc}$

$\sqrt{bc}+k\sqrt{a} = \frac{1}{2}(k^2+a-b-c) \in\mathbb{Q}$

Then from my post here we can deduce that $\sqrt{a}\in\mathbb{Q}$ . For convenience I'll prove it here:

$k\sqrt{a}+\sqrt{bc}\in\mathbb{Q} \Rightarrow k\sqrt{a}-\sqrt{bc} = \frac{k^2a-bc}{k\sqrt{a}+\sqrt{bc}} \in\mathbb{Q}$

Therefore

$\sqrt{a} = \frac{1}{2k} \big((k\sqrt{a}+\sqrt{bc})+(k\sqrt{a}-\sqrt{bc})\big) \in\mathbb{Q}$

and we're done (under the assumption that $k\ne 0$ . Otherwise $a=b=c=0$ and the result is trivial)

**JRichardson1729** · October 18th 2010, 08:23 AM

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.

**Unbeatable0** · October 18th 2010, 09:05 AM

Originally Posted by JRichardson1729

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 $a,b,c$ are rational numbers and $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is also rational,

then all those square roots are rational themselves.

Applied to your problem - we're given that $p,q,r$ are integers

and $\sqrt{p}+\sqrt{q}+\sqrt{r}$ is also an integer. Since being

an integer is just a particular case of being rational, the above argument shows

that all $\sqrt{p},\:\sqrt{q},\:\sqrt{r}$ are rational. From here,

can you deduce that $\sqrt{p},\:\sqrt{q},\:\sqrt{r}$ 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)

**JRichardson1729** · October 19th 2010, 08:11 AM

Originally Posted by Unbeatable0

Therefore

$\sqrt{a} = \frac{1}{2k} \big((k\sqrt{a}+\sqrt{bc})+(k\sqrt{a}-\sqrt{bc})\big) \in\mathbb{Q}$

This may be another stupid question, but I do not understand how you have produced this equation. I know that if you simplify it, it simply becomes $\sqrt{a}=\sqrt{a}$ , but I don't understand how you find this out from the previous parts of the proof.

**Unbeatable0** · October 19th 2010, 08:25 AM

Originally Posted by JRichardson1729

This may be another stupid question, but I do not understand how you have produced this equation. I know that if you simplify it, it simply becomes $\sqrt{a}=\sqrt{a}$ , but I don't understand how you find this out from the previous parts of the proof.

It came from noticing that if $\sqrt{a}+\sqrt{b}$ is rational then so is $\sqrt{a}-\sqrt{b}$ (when $a,b$ are rational).

But then, since we know that sum of rational numbers is rational we conclude that $2\sqrt{a}$ is rational.

Dividing by 2 (since it's rational) yields $\sqrt{a} \in\mathbb{Q}$ . The same logic was applied

in the above proof, but for $k\sqrt{a}$ and $\sqrt{bc}$ - we see that it gives us

$2k\sqrt{a}\in\mathbb{Q}$ and we can divide out the $2k$ . What I did was just

to squeeze all these steps together.

**JRichardson1729** · October 25th 2010, 08:17 AM

Originally Posted by Unbeatable0

From here,

can you deduce that $\sqrt{p},\:\sqrt{q},\:\sqrt{r}$ 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)

I cannot seem to prove this. Could you please help me with a proof? Thank you for all of your time and help.

**Unbeatable0** · October 25th 2010, 08:25 AM

Originally Posted by JRichardson1729

I cannot seem to prove this. Could you please help me with a proof? Thank you for all of your time and help.

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 $x^2-a$ . From this, if $\sqrt{a}$ is rational then it's integer (assuming $a$ is integer), which is what we want.

**JRichardson1729** · October 25th 2010, 08:39 AM

Originally Posted by Unbeatable0

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 $x^2-a$ . From this, if $\sqrt{a}$ is rational then it's integer (assuming $a$ is integer), which is what we want.

Again, my lack of mathematical knowledge is limiting my ability to adapt the infinite descent proof of the irrationality of $\sqrt{2}$ . I can get as far as $a^2=xb^2$ , but then I cannot continue.

**Unbeatable0** · October 25th 2010, 08:49 AM

Originally Posted by JRichardson1729

Again, my lack of mathematical knowledge is limiting my ability to adapt the infinite descent proof of the irrationality of $\sqrt{2}$ . I can get as far as $a^2=xb^2$ , but then I cannot continue.

Then I suggest that you go by the second way I proposed. Anyway, for the first way:

Say $\sqrt{x} = \frac{a}{b}$ , with $a,b$ coprime . Then $a^2,b^2$ are coprime,

so that $x = \frac{a^2}{b^2}$ can be an integer iff $b^2=1$ . That is, $b=1$ .

**JRichardson1729** · October 25th 2010, 08:59 AM

Sorry, I don't understand what the Rational Roots theorem states.

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