Question: I have this simple looking problem, but I am stumped. It isn't for homework, it's just for fun.

Solve for $x$ and $y$, $(\sqrt x+\sqrt 3)^2=y+6\sqrt 2$, where $x$ and $y$ are positive integers.

I get $x=6$ and $y=9$ from Wolfram, but I'm not sure about steps. Can anyone help?

So far I have $x+2 \sqrt {3x} + 3 = y + 6\sqrt 2$

Answer: 
No! There is no reason why $\pi^\*TB$ is a subbundle of $TE$ -- this is precisely why you need a connexion.

Recall $\pi^\*TB$ is actually constructed as $E\times\_B TB$:
$$
\pi^\*TB=\{(e,(p,v))\in E\times TB\mid \pi(e)=\operatorname{proj}\_{TB\to B}(p,v)=p\}
$$
where $\operatorname{proj}\_{TB\to B}$ is the usual projection $TB\to B$.
By definition of the pullback, $\pi^\*B=\{(x,y,v):y\in\pi^{-1}(x), v\in T\_xB\}$ therefore it is not a subbundle of $TE$.