$100-50\sqrt3$ to be a perfect square

Question

If we want $100-50\sqrt3$ to be a perfect square $(a+b)^2=a^2+2ab+b^2$, most likely it is the case that $-25\sqrt3$ corresponds to $ab$. I can't approach the problem further. Can you give me a hint?

Answer 1

Writing the square as $(a-b\sqrt 3)^2$, you haven't many simple choices: from $ab=25$, you deduce that either $\{a,b\}=\{1,25\}$ or $a=b=5$.

Answer 2

As you noted, we have: $$ab=25\sqrt{3}\leftrightarrow a=\frac{25\sqrt3}{b}$$ From here, we substitute and obtain: $$\left(\frac{25\sqrt3}{b}\right)^2+b^2=100$$ Let $t=b^2$, we have a quadratic equation: $$t^2-100t-1875=0\leftrightarrow t=\frac{100\pm\sqrt{100^2-4\cdot1875}}{2}=75 \vee 25$$ In conclusion there are two possibilities: $$b=5 \land a=5\sqrt{3} \: \; \; \vee \;\;\; b=5\sqrt{3} \land a=5$$

Answer 3

looking at what the answer is, it can be deduced that a+b is in the form x+y√z. if you square that you get x^2+2xy√z+zy^2. You then match up 100 with x^2+zy^2 and -5√3 with 2xy√z, to get x^2+z*y^2 =100 and 2xy√z=-5√3. From the second equation it becomes clear that z is equal to 3, so you can replace z with 3 in both equations: x^2+3*y^2=100 and 2xy√3=-5√3 --> 2xy=-5. Now your equations are a bit more friendly. However, I'm not really sure the math will turn out too different this is just what I thought of first.

Answer 4

To simplify the question slightly, note that $\sqrt{100 - 50 \sqrt 3} = \sqrt{25(4-2 \sqrt 3)} = 5 \sqrt{4 - 2 \sqrt 3}$.

Therefore if $4 - 2 \sqrt 3$ can be written in the form $(a-b\sqrt3)^2$, $a^2+3b^2 = 4$ and $-2ab\sqrt3= -2\sqrt{3} \Rightarrow ab = 1$.

The rest is just standard algebra:

$$(1/b)^2 + 3b^2 = 4 \Rightarrow 1 + 3b^4 = 4b^2 \Rightarrow 3b^4-4b^2+1 = 0 \Rightarrow (3b^2-1)(b^2-1)=0$$

$$b^2 = \frac{1}{3}, 1$$ $$b = ±\frac{1}{\sqrt 3}, ±1$$ $$a = \mp \sqrt3, \mp 1$$

The solutions $b = \frac{1}{\sqrt 3}, a = -\sqrt{3}$ and $b = 1, a = -1$ are actually equivalent. In addition, only $a = -1, b = 1$ fits because the number you are looking for is positive.

Therefore you have $5 \sqrt{(-1 + \sqrt 3)^2} = 5(\sqrt 3 - 1) = 5 \sqrt 3 - 5$.