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Can anyone explain to me how/why this works (hopefully in mostly layman's terms)?
It seems pretty magical to me at the moment.

$${{(a+b)^2\over4} - {(a-b)^2\over4}} = a b.$$

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    $\begingroup$ The effect is even bigger if you give names to quantities: $(a+b)/2$ is the average and $|a-b|/2$ the half-distance; so if you subtract the square of the half distance from the square of the average, voila, you get the product! You can then impress your friends more by saying: now take the difference and the sum, square them and subtract the squared difference from the squared sum. You will get the product of their doubles! And people call elementary algebra boring... $\endgroup$ – guest Nov 10 '14 at 3:11
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    $\begingroup$ @guest: . . . only in comparison to real Algebra. $\endgroup$ – imallett Nov 10 '14 at 5:45
  • $\begingroup$ @imallett Indeed! But I was referring to the general population, rather than people in the exact and mathematical sciences. Abstract algebra is not easily accessible or meaningful to people with only a high school education; in contrast, every high schooler has practiced some elementary algebra. If more secondary school teachers would present examples of its elegance and interesting consequences like the one in the OP, I think more students would become interested in mathematics. $\endgroup$ – guest Nov 10 '14 at 6:04
  • $\begingroup$ @guest, quite. I have a post that touches on this. It's a pity Algebra isn't taught at lower levels anymore--but, as an educator, I can always blow peoples' minds by explaining that "Algebra" actually means a whole lot more. $\endgroup$ – imallett Nov 10 '14 at 6:11
  • $\begingroup$ Which positive integers are boring? If there are any, then there must be a smallest boring number. Isn't that interesting! (little joke) $\endgroup$ – Walter Mitty Nov 10 '14 at 11:30
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Try a geometric argument          

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    $\begingroup$ That's beautiful. $\endgroup$ – Sam Washburn Nov 10 '14 at 19:20
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    $\begingroup$ To not write anything use \mathbb{} in mahjax- it is blank characters, often used for this purpose. $\endgroup$ – Joao Nov 11 '14 at 2:05
  • $\begingroup$ @Joao:   works well, too. $\endgroup$ – Ilmari Karonen Nov 11 '14 at 19:14
  • $\begingroup$ @IlmariKaronen Cool. I'll use that next time. $\endgroup$ – Joao Nov 12 '14 at 0:39
  • $\begingroup$ -1 Where did the quarters go? $\endgroup$ – Jostein Trondal Dec 8 '14 at 9:35
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Expanding the squared terms gives \begin{equation} \frac{(a + b)^2}{4} - \frac{(a - b)^2}{4} = \frac{a^2 + 2ab + b^2}{4} - \frac{a^2 - 2ab + b^2}{4} = \frac{4ab}{4} = ab. \end{equation}

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In addition to the direct derivations already shown, your magical equation is closely related to the formula $$x^2 - y^2 = (x + y)(x - y).$$ Just set $a = x + y$ and $b = x - y.$ Then $\frac{a+b}2 = x$ and $\frac{a-b}2 = y,$ so $x^2 = \frac{(a+b)^2}4$ and $y^2 = \frac{(a-b)^2}4.$ Use these facts to replace $x^2,\ y^2,\ x + y,$ and $x - y$ in the equation above and you will have derived your magical equation in $a$ and $b.$

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Your expression is equal to $\dfrac{a^2+2ab+b^2-a^2+2ab-b^2}{4}=\dfrac{4ab}{4}=ab$

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Try writing the fractions $\frac{(a+b)^2}{4}$ and $\frac{(a-b)^2}{4}$ as one, and expanding the brackets. See what happens then.

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An underrated way to show identities is to use the fact that if $A-B=0$ then $A=B$:

$$\begin{align*} &\ \frac14(a+b)^2-\frac14(a-b)^2-ab\\[3mm] =&\ \frac14(a^2+2ab+b^2)-\frac14(a^2-2ab+b^2)-ab\\[3mm] =&\ \frac14a^2-\frac14a^2+\frac12ab+\frac12ab+\frac14b^2-\frac14b^2-ab\\[3mm] =&\ ab-ab\\[3mm] =&\ 0 \end{align*}$$

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Here's a really detailed step by step proof, suitable for elementary or secondary school students:

$$ \begin{align} \frac{(a+b)^2}{4}-\frac{(a-b)^2}{4}&=ab\\\\(a+b)^2 - (a-b)^2&=4ab\\\\a^2 + 2ab + b^2 - (a^2 - 2ab + b^2)&=4ab\\\\a^2 + 2ab + b^2 - a^2 + 2ab - b^2 &= 4ab\\\\(a^2 - a^2) + (b^2 - b^2) +2ab + 2ab &= 4ab\\\\4ab &= 4ab \end{align} $$

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    $\begingroup$ Eww. Trig-style proof. You should really go the other way around, or down one side and up the other. $\endgroup$ – Justin Nov 10 '14 at 5:07
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    $\begingroup$ You're starting by assuming equality... That's not good math. You could add some more steps into yoknapatawpha's proof above to get a really nice, simple proof, but what you have right now is not a proof. $\endgroup$ – Johanna Nov 10 '14 at 6:00
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    $\begingroup$ @Johanna Huh? Of course it's a good proof. They's not starting off by assuming equality, they're showing that the first equation is equivalent to the last. $\endgroup$ – Jack M Nov 10 '14 at 10:12
  • $\begingroup$ @JackM Maybe DrC is doing what you say he (or she) is, but that's not clear. Johanna's right. This proof would be OK if between each equation it said explicitly "which is equivalent to". Without that, it really does seem to reason from what its trying to prove. $\endgroup$ – Ethan Bolker Nov 11 '14 at 0:41
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    $\begingroup$ @JackM: A sequence of formulas unaccompanied by any indication of how they're claimed to relate to each other is never a good proof, no matter if the formulas in the sequence could possibly appear in an actual proof. $\endgroup$ – hmakholm left over Monica Nov 11 '14 at 10:14

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