Given $a + \sqrt{b}$ with positive integer $a,b$, find $a$ and $b$? Suppose I had $n = a + \sqrt{b}$ as a decimal of arbitrary precision, but didn't know $a$ or $b$, except that they are positive integers. 
If I had just $\sqrt{b}$, I could just square it and end up with something very close to an integer, so I'd have $b$.
If I take $n^2 - 2an$, I get $(a^2 + b + 2a\sqrt{b}) - (2a^2 + 2a\sqrt{b}) = b - a^2$ which will be an integer, but I don't have $a$...
Is there some way to take the sum of the integral and decimal portion of $n$ and do some integer voodoo there?
Thanks for any ideas.
 A: You can calculate its continued fraction. Being a quadratic irrational, its continued fraction will be periodic. Once you have the period, you can work out $a$ and $b$ exactly. 
EDIT: Probably a better method is to calculate the square of your number and then use an algorithm like PSLQ to look for a vanishing integer-linear combination of $1$, the number, and its square. 
A: Using that $\sqrt{b}=\lfloor\sqrt{b}\rfloor+\{\sqrt{b}\}$, we get that $\{\sqrt{b}\}$ satisfies the following:
$$\{\sqrt{b}\}^2+2\{\sqrt{b}\}\lfloor\sqrt{b}\rfloor+\lfloor\sqrt{b}\rfloor^2=b$$
and in particular it satisfies a relation of the form $\{\sqrt{b}\}^2+m\{\sqrt{b}\}=k$ for some positive integers $m,k$ (you can just check all $m$ up to $2n$ to find the solution). By irrationality of $\sqrt{b}$ (if it's not irrational, then the problem is unsolvable anyway), there is only one solution and once you have it, $\lfloor\sqrt{b}\rfloor$ must be $\frac{m}{2}$ and $k$'ll be $b-\lfloor\sqrt{b}\rfloor^2$, so $b$'ll be $k+\frac{m^2}{4}$.
A: Clearly $a \leq n$ and $b \leq n^2$ so you have only got finitely many possibilities to try. Compute the value of $x+\sqrt{y}$ to 100 decimals (say) for all positive integers $x,y$ such that $x\leq n$ and $y \leq n^2$. (You only need an upper bound for $n$ and $n^2$ here.) Also compute $n$ to 100 decimals. If you are lucky there is now only one combination of $x,y$ such that $x+\sqrt{y}$ could be equal to $n$. Then $a=x$ and $b=y$. Otherwise you double the number of digits you compute. Repeat as long as necessary. 
A: Assuming that the integer part of $n$ is $N$, then you need to try the following 

for $k=1$ to $N$ 
store smallest fractional part of ${(N-k)}^2$ 

(note if the smallest fractional part is $>0$ for example, $0.001$, then you would store that, and if it is close to $1$, for example, $0.998$, you would store $1-0.998=0.002$) 
The $k$ that gave you the smallest value is your $a$.
