# Interesting question about irrational numbers

Find all solutions in un-ordered integers $(a,b)$ to $7-a-b=2\sqrt{10}-2\sqrt{ab}$. It would appear that the only solution to this is $a=2, b=5$. But how to prove this rigorously? Do irrational numbers have a unique (after decimal point) representation? i.e $-------...----.a_1a_2a_3a_4 \cdots$ and $-------...---.a_1a_2a_3a_4 \cdots$

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Why are there two questions? I think thta a unique question for each title can help in obtaining beter answers. Regards. –  awllower Apr 28 '13 at 5:14
The second question is almost the same, perhaps a bit more general as we want to know if $\sqrt{10}-\sqrt{ab}$ can equal any integer other than $0$. –  John Marty Apr 28 '13 at 5:20
Since $7-a-b$ is an integer, $2\sqrt{10}-2\sqrt{ab}$ must also be an integer, and $\sqrt{10}$ is irrational. That should help. –  Peter Grill Apr 28 '13 at 5:22
$a=5, b=2$ is also a solution. –  MJD Apr 28 '13 at 5:29
Yes you're right, I should have said un-ordered pairs. I have edited it appropriately. Thanks. –  John Marty Apr 28 '13 at 5:31
Squaring both sides you get $(7 - a - b)^2 - 40 - 4ab = -8\sqrt{10ab}$. Since $10ab$ is an integer, and the square root of an integer is always an integer or irrational, we get that $\sqrt{10ab}$ is an integer. This shows that $ab = 10r^2$ for some $r \in \mathbb{Z}$. Then, the original equation implies $7 -a - b = 2\sqrt{10}(1 - |r|)$. Since $\sqrt{10}$ is irrational and everything else in this last equation is an integer, we obtain $1 - |r| = 0$. Thus, $r = +1$ or $-1$. Hence, $ab = 10$. This restricts our choices for $a, b$ because they have to be divisors of $10$ so can only be from the set $\{1,-1,2,-2,5,-5, 10, -10\}$. You can take it from there.