Is this relationship already known? 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
 A: We have $$b=a+1$$
This implies
$$b^2=(a+1)^2=a^2+2a+1=a^2+a+a+1=a^2+a+b$$
A: Not really. Can be easily derived:
$$b^2 = a^2 + a + b$$
$$\iff b^2 -a^2 =  a + b$$
$$\iff (b -a)(b+a) =  a + b$$
$$\iff b -a =  1 \lor a + b = 0$$
A: 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.  
A: 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 $b=a+1$.
