Chinese New Year Equation 2016

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 $a=22$.

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 $$(a-21)^2 = \frac{6257}{16}-(18n+\frac{15}{4})^2$$

$$\color{red}{\text{Happy 2016!}}$$