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March 9th, 2017, 12:32 AM   #1
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Red face Need help.

Is 3+2√2 a perfect square?

Please help.
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March 9th, 2017, 12:38 AM   #2
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Let’s check if it is…

$\begin{equation} \begin{split}3+2 \sqrt{2}&=3+2 \cdot \sqrt{2} \cdot 1\\&=2+2 \cdot \sqrt{2}\cdot 1+1\\&=(\sqrt{2})^2+2 \cdot \sqrt{2} \cdot 1+1^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \bigg\{\because a^2+2ab+b^2=(a+b)^2\bigg\} \\&=(\sqrt{2}+1)^2 \end{split}\end{equation} \tag*{}$

Yes, it is a perfect square!
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Last edited by deesuwalka; March 9th, 2017 at 12:43 AM.
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March 9th, 2017, 08:59 AM   #3
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Hmmm....isn't a "perfect square" the square of an integer?
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March 11th, 2017, 12:59 PM   #4
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That is my understanding also...
A perfect square is the the squared value of an integer, the result of an integer multiplied by itself.
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March 11th, 2017, 01:48 PM   #5
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I think the definition varies on context..

Perfect Squares | What is a Perfect Square | Math@TutorVista.com

https://en.m.wikipedia.org/wiki/Square_number

If we just say $3 + 2\sqrt{2}$ is a number, any number, then it can be shown to have a perfect square.

Square number and perfect square are probably better terms to distinguish the two.
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March 11th, 2017, 01:56 PM   #6
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I like the tricks used by deesuwalka

I guess we are relaxing the condition on what a perfect square is.

Is $ \ \ \ \ a + b \sqrt{c} \ \ \ \ $ a perfect square?

I would like to see an algebraic demonstration , the way deesuwalka showed. If possible.

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March 11th, 2017, 03:11 PM   #7
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Deesuwalka does have a neat analysis.

I agree with denis that I have always understood the definition of a perfect square to be an integer with a square root that is also an integer.

But quite clearly that definition cannot be intended here. So context asks for a different definition (and of course we do not know what the original problem says). The definition implied here seems to be whether a number of the form

$a + b\sqrt{c},\ a,\ b,\ c \in \mathbb Z_+,\ and\ c > 1 \implies \exists\ p,\ q,\ r \in \mathbb Z$

$such\ that\ \left ( p + q\sqrt{r} \right )^2 = a + b\sqrt{c}.$

It is easy to show that

$a = p^2 + q^2r\ and\ b = 2pq\ and\ c = r \iff \left ( p + q\sqrt{r} \right )^2 = a + b\sqrt{c}.$

Thus we can show that for p, q, and r to be integers, c must be an integer (true by hypothesis). Also, p and q be integers that satisfy

$p^2 + q^2c = a\ and\ b = 2pq \implies p = \dfrac{b}{2q} \implies$

$\dfrac{b^2}{4q^2} + q^2c = a \implies a = \dfrac{b^2 + 4q^4c}{4q^2}.$

So deesuwalka's approach is very clever, but will not work for every case because the "perfectness" does not apply to every case.

Note that in this problem $p = 1 = q\ and\ r = 2.$

And c = 2. That checks.

And $b = 2pq = 2 * 1 * 1 = 2.$ That checks.

And $a = \dfrac{2^2 + 4 * 1^4 * 2}{4 * 1^2} = \dfrac{4 + 8}{4 * 1} = 3.$ That checks.
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March 11th, 2017, 04:05 PM   #8
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...plus will create havoc with the sum of 1st n squares formula
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