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 March 9th, 2017, 01:32 AM #1 Newbie   Joined: Mar 2017 From: US Posts: 1 Thanks: 0 Need help. Is 3+2√2 a perfect square? Please help.
 March 9th, 2017, 01:38 AM #2 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 Let’s check if it is… $$$\begin{split}3+2 \sqrt{2}&=3+2 \cdot \sqrt{2} \cdot 1\\&=2+2 \cdot \sqrt{2}\cdot 1+1\\&=(\sqrt{2})^2+2 \cdot \sqrt{2} \cdot 1+1^2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \bigg\{\because a^2+2ab+b^2=(a+b)^2\bigg\} \\&=(\sqrt{2}+1)^2 \end{split}$$ \tag*{}$ Yes, it is a perfect square! Thanks from agentredlum and JeffM1 Last edited by deesuwalka; March 9th, 2017 at 01:43 AM.
 March 9th, 2017, 09:59 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,399 Thanks: 731 Hmmm....isn't a "perfect square" the square of an integer?
 March 11th, 2017, 01:59 PM #4 Newbie   Joined: Feb 2017 From: Omaha, Nebraska Posts: 8 Thanks: 0 Math Focus: Algebra and Trigonometry That is my understanding also... A perfect square is the the squared value of an integer, the result of an integer multiplied by itself.
 March 11th, 2017, 02:48 PM #5 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,476 Thanks: 495 Math Focus: Yet to find out. 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.
 March 11th, 2017, 02:56 PM #6 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 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. Thanks from deesuwalka
 March 11th, 2017, 04:11 PM #7 Senior Member   Joined: May 2016 From: USA Posts: 883 Thanks: 353 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. Thanks from agentredlum, Joppy and deesuwalka
 March 11th, 2017, 05:05 PM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,399 Thanks: 731 ...plus will create havoc with the sum of 1st n squares formula

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