Chinese New Year Equation 2016 In the spirit of Chinese New Year, here's a problem to commemorate the year. 

$\color{black}{\text{Solve the following equation for positive integers $a$ and $b$:}}$
  $$\color{red}{a^2+b^2+(a+8)^2+(b+8)^2=100a+b}$$


Edit 1 
As this question has been put on hold and the OP asked to improve the question by providing additional context", please find below some additional details. 
Having read about Lagrange's four-square theorem and after some work, I noticed an interesting four-square combination for the year. Hence I thought I would formulate it as a problem - which turns out to be a diophantine equation - to see what different approaches there might be. A straightforward approach would be to rearrange the terms into a the standard circle formula
$(a-21)^2+\left(b+\frac {15}4\right)^2=\left(\frac{\sqrt{6257}}4\right)^2\approx19.77^2$ and then testing integer values of $a$ within $21\pm 19.77$, but there should be other more interesting approaches.  
Hopefully the explanation above will be sufficient for the question to be reopened. Moderators - please advise if there is additional context required. 
Thanks.

Edit 2
Thanks for the nice answers by Hagen and Daniel. The interesting point to note here is that:

$$\color{red}{20}^2+\color{red}{16}^2+(\color{red}{20}+8)^2+(\color{red}{16}+8)^2=\color{red}{2016}$$

$\color{black}{\text{To all readers who celebrate, a very Happy Chinese New Year!!}}$

Note: This question has been put on hold.  If you find it interesting or useful, please vote to reopen it (by clicking on "reopen" at the bottom of this post), so that others can post their solutions. Thanks!
 A: The difference of the two sides (which must be $0$) equals
$$\tag12(a-21)^2 +2(b-11)^2+59b-996\ge 59b-996$$
so that $b\le 16$. Also, the left hand side of $(1)$ is $\equiv b\pmod 2$, hence $b$ is even, $b=2c$ with $1\le c\le 8$. Substituting and dividing by $2$ we arrive at
$$ 0=(a-21)^2+\underbrace{4c^2+15c-377}_{-358\,\ldots\,-1}$$
and check for which $c\in\{1,\ldots,8\}$ the number $377-4c^2-15c$ happens to be a perfect square. The only case is $c=8$ (so $b=16$) and leads to $a-21=\pm1$, i.e., $a=20$ or $a=22$.
A: First write it as $$(a-21)^2 +(b+\frac{15}{4})^2= \frac{6257}{16} = 1+\frac{79^2}{4^2}$$
$$\Rightarrow (a-21)^2-1 = \frac{79^2}{4^2}-(b+\frac{15}{4})^2$$
Now use difference of squares to get
$$(a-22)(a-20) = \bigg( \frac{4b+94}{4}\bigg)\bigg( \frac{64-4b}{4}\bigg)$$
We see we get corresponding positive integer solutions in $b$ for $a = 22$ and $a = 20$(since we only need one of the factors on the RHS to be $0$), namely $b = 16$. 
Else we need RHS to be an integer (since the LHS clearly is), which will happen only when$$(4b+94)(64-4b) \equiv 0 \pmod{16} $$ $$\Rightarrow b = 18n$$
However it is easy to check that we get no new solutions from this. To see this write it as $$(a-21)^2 = \frac{6257}{16}-(18n+\frac{15}{4})^2$$
$$\color{red}{\text{Happy 2016!}}$$
