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Choose any number, $x$: say, $x = 876$ (you can pick any $n$ digit number)

Now, square the number -> $876 * 876 = 767376$

But now, If I ask you the square of $ x + 1$ --> $876 + 1 = 877$. You can't solve it mentally (in most of the cases).

But instead, you can, take $x^2$ and add $ x + \{x + 1\}$ to it. Which gives --> $ x^2 + x + \{x + 1\} = 767376 + 876 + 877 = 769129 $ which interestingly is actually the square of 877!

So, this all comes down to one thing, have I discovered some new mathematical logic?

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    $\begingroup$ You seem to have 'discovered' that $(x+1)(x+1)=x^2+2x+1$ $\endgroup$ – user2520938 Jul 31 '15 at 20:14
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    $\begingroup$ Happy: it is always fun discovering interesting things. Sad: you probably can't publish this result. $\endgroup$ – Umberto P. Jul 31 '15 at 20:17
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    $\begingroup$ @Anoneemus Happy: thinking deeply about something sometimes makes you loose sight of the bigger picture, but that doesn't mean it's not fun! Stuff happens $\endgroup$ – user2520938 Jul 31 '15 at 20:17
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    $\begingroup$ I don't want to run down your discovery, which is legitimate mathematics, and you should be pleased with yourself for discovering it. But it is not publishable anywhere except perhaps on your blog, because it was discovered in prehistoric times. $\endgroup$ – MJD Jul 31 '15 at 20:22
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    $\begingroup$ @Anoneemus I think your insight is fun, thanks for sharing it. Seems to me it often can be faster and easier to remember to add $x$ and its successor than to remember to add $2x+1$. $\endgroup$ – Bob Pego Jul 31 '15 at 21:03
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You will notice in your general case you have $x^2+2x+1$, which is precisely the binomial expansion of $(x+1)^2$. Thus essentially you derived the logic behind the expansion of binomials.

So the result makes sense, if you want the square of some $x$ plus one you do indeed want $(x+1)^2$ or $x^2+2x+1$. This isn't a new result, but the exploration which led you to it is a very good mental exercise.

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