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I like math because it's a puzzle to me, but am really not very good at it. But I figured out the relationship below myself. Just curious, is this already pretty common knowledge? Kind of proud of myself for figuring it out, but my son who's getting math minor had never heard of it. Apologies in advance for any poor explanation.

$b^2 = a^2 + a + b$ for positive integers where $b - a = 1$

e.g. $2^2 = 4$

$3^2 = 9$

$4 + (2 + 3) = 9$

$4^2 = 16$

$9 + (3 + 4) = 16$

and so on.

Edit in response to @fleablood comment: Where b = a + 2 Even numbers... 2^2 = 4 -> a^2 = a2 4^2 = 16 -> b^2 = b2 16 - 4 = 12 -> b2 - a2 = c (a + 1) * b = 3 * 4 = c Odd Numbers... 3^2 = 9 -> a^2 = a2 5^2 = 25 -> b^2 = b2 25 - 9 = 16 -> b2 - a2 = c (a + 1) * b = 4 * 5 = 20 c != 20

recurvata
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    Your chance of stumbling upon something novel with such basic arithmetic is pretty remote. But I commend your enthusiasm. – Gregory Grant Feb 01 '16 at 23:23
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    No doubt! But it's a step forward for me. – recurvata Feb 01 '16 at 23:36
  • Discovering something that is both new and interesting is really hard. But trying something, noticing a pattern, and formalizing it is (I think) how a lot of math gets done, so even if it's not new math, it's still math. Good for you! – David K Feb 01 '16 at 23:54

4 Answers4

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We have $$b=a+1$$

This implies

$$b^2=(a+1)^2=a^2+2a+1=a^2+a+a+1=a^2+a+b$$

Peter
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  • Thanks Peter & @adjan. Giving the answer credit to Peter because he was first, but both are good answers. Except not sure what double headed arrows mean. Anyways, now working on a formula for a^2 and (a + 2) ^2. So far, seems to be a different relationship depending on if the integers are odd or even. – recurvata Feb 01 '16 at 23:42
  • Not really. $(a + 2)^2 = a^2 + 4a + 4 = a^2 + (2(a+1) + 1) + (2a + 1)$ – fleablood Feb 01 '16 at 23:46
  • If $b = a+2$ then $b^2=(a+2)^2=a^2+4a+4=a^2+2a+2(a+2)=a^2+2a+2b$. – David K Feb 01 '16 at 23:51
  • double arrows mean that if one of the statements is true it implies the other one is true and likewise if the other one is true the first one is. We call it "if and only if". It's an if statement that works both ways. $a$ is even => $a + 1$ is odd. That's a one way implicatation $a + 1$ is odd => $a$ is even. That's another one-way implication. So $a+1$ is odd <=> $a$ is even. "a plus one is odd if and only if a is even". – fleablood Feb 02 '16 at 00:00
  • Ok, thanks. Edited original post. Like I said, not that good at math. Come from more of a programming background. Kind of wondering if different ways would be faster than others. – recurvata Feb 02 '16 at 00:16
  • @recurvata In general, $b^2-a^2=(b+a)(b-a)$. You can rearrange that to $b^2=a^2+a(b-a)+b(b-a)$. So if $b=a+x$, then $b^2=a^2+xa+xb$. – f'' Feb 02 '16 at 04:09
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Not really. Can be easily derived:

$$b^2 = a^2 + a + b$$

$$\iff b^2 -a^2 = a + b$$

$$\iff (b -a)(b+a) = a + b$$

$$\iff b -a = 1 \lor a + b = 0$$

adjan
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This is another way of stating that $n^2 = 1 + 3 + 5 + ...$; the sum of the first n-1 odd numbers.

$a - b = 1$ means $a = b + 1$ means $a + b = 2b + 1$ means $a + b$ is the (b-1)-th odd number.

So $a^2 = b^2 + a + b$ is the same as saying $(b + 1)^2 = b^2 + (b + 1) + b = b^2 + 2b + 1$; which can be shown by expansion: $(b + 1)(b+1) = b(b+1) + (b + 1) = b^2 + b +b + 1 = b^2 + 2b + 1$.

Expanding it is $a^2 = a + b + b + (b-1) + (b-1) + .... + 1 + 1$ = $a + 2(a-1) + 2(a - 2) + .... + 4 + 2$.

... which is another way of saying $a^2 = 1 + (1+2) + (2+3) + .... + ((a - 2) + (a-1)) + ((a-1) + a)$.

... which is another way of saying $a^2 = 1 + 3 + 5 + ... [2(a-1) + 1]$

... which is another way of saying $a^2 = (a -1)^2 + 2(a-1) + 1$

I remember being surprised when I first discovered it, too.

fleablood
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  • I get a little lost as you get deeper into it, but I really like your first statement. That is a really elegant and understandable way of stating it. This is one of the things I like about math, and why I wish I was better at it. It's seeing the relationships between things at some kind of fundamental level. It's an awesome accomplishment that math, an intellectual construct, can actually model the universe. – recurvata Feb 02 '16 at 00:31
  • Another way of looking at it is $(a + 1)^2 - a^2 = 2a + 1$ which is always odd and an odd that increases by two each time . $4-1 =3; 9-4 =5; 16-9 = 7; 25 - 16 = 9;$ etc. This all comes for $(a + 1) ^2 = a^2 + (2a + 1)$. – fleablood Feb 02 '16 at 01:13
  • Three things I'm proud of of discovering on my own in Jr. High: 1) $1 + .... + n = n(n+1)/2$. 2)$1 + 3 + ... + 2n - 1 = n^2$ which I figured went $1 + 3 ... = x$ then $2x = (1 + 3 + ... + 2n-1) + (2n-1 + 2n -3 + ....+ 3 + 1) = (1 + 2n -1) + (3 + 2n - 3) + ... (2n -3 + 3) + (2n -1 + 1) = 2n + 2n + ...+2n = n*2n = 2n^2$. – fleablood Feb 02 '16 at 01:21
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I would prove it this way: \begin{align} b^2&=a^2+a+b&\iff\\ b^2-b&=a^2+a&\iff\\ (b-1)(b)&=(a)(a+1) \end{align} which is clearly true if $b=a+1$.

Akiva Weinberger
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