Solved Challenge Solving equation

[anemone]

Find a real number $x$ that satisfy the following equation:

$x = \sqrt{(x-667)(x-736)}+\sqrt{(x-736)(x-928)}+\sqrt{(x-928)(x-667)}$

 

[Opalg]

Spoiler

Optimistic guess: suppose that $x-667$, $x-736$ and $x-928$ are all perfect squares, say $x-667 = a^2$, $x-736 = b^2$ and $x-928 = c^2$. Then $a^2 - b^2 = 736 - 667 = 69$. But if $a$ and $b$ are integers with $(a+b)(a-b) = 69$ then either $a+b=23$ and $a-b = 3$ or $a+b=69$ and $a-b = 1$. In the first of those cases, $a=13$ and $b=10$. Then $x-667 = 13^2 = 169$, so that $x= 836$. But if $x = 836$ then $c^2 = 836 - 928$, which is negative. So that solution does not work.

I then tried the other possible case, with $a+b=69$ and $a-b = 1$, but that did not lead to a solution either. So it looked as though there was not going to be a solution in integers.

But then the fact that $c^2$ turned out to be negative in my first attempt at a solution made me think about the possibility that $x-667$, $x-736$ and $x-928$ might all be negatives of squares. In that case, my first solution would have to be adjusted, to make $x-667 = -100$ so that $x=567$. Then $x-736 = -169 = -13^2$ and $x-928 = -361 = -19^2$.

Finally, if $x = 567$ then $$\begin{aligned}\sqrt{(x-667)(x-736)}+\sqrt{(x-736)(x-928)}+\sqrt{(x-928)(x-667)} &= \sqrt{(-100)(-169)} + \sqrt{(-169)(-361)} + \sqrt{(-361)(-100)} \\ &= 10*13 + 13*19 + 19*10 \\ &= 130+247+190 \\ &= 567.\end{aligned}$$

 

[Theia]

Spoiler

If I see correctly, there should be 2 distinct solutions over real numbers for the equation, one of which around \( x \approx 1184.13 \). Does the finding of this include to the challenge too?

 

[anemone]

Hi Theia , as long as you find one real $x$ value that satisfies the given equation, then that $x$ will be an answer to the problem.

 

[Theia]

Spoiler

We can write the original equation in the form
\[x = \sqrt{a} + \sqrt{b} + \sqrt{c},\]
where
\[\begin{align*}
a &= (x - 667)(x - 736) \\
b &= (x - 736)(x - 928) \\
c &= (x - 928)(x - 667).
\end{align*} \]
By solving the $\sqrt{c}$ from the equation and squaring one obtains
\[c = x^2 - 2x\sqrt{b} - 2x\sqrt{a} + b + 2\sqrt{a}\sqrt{b} + a. \]
Let's now move all terms with $\sqrt{b}$ dependence to left hand side and others to right hand side:
\[2\sqrt{b} \left( \sqrt{a} - x \right) = -x^2 + 2x\sqrt{a} + c - b - a.\]
After squaring only $\sqrt{a}$ is left. Collecting those terms and squaring one can write \[\begin{align*}
& x^4 - 4x^3\sqrt{a} - 2cx^2 - 2bx^2 + 6ax^2 + 4cx\sqrt{a} + 4bx\sqrt{a} - 4ax\sqrt{a} \\
& \quad + c^2 - 2bc - 2ac + b^2 - 2ab + a^2 = 0 \qquad \Leftrightarrow \end{align*} \]
\[\begin{align*}
4x\sqrt{a} \left( x^2 - c - b + a \right) = &\ x^4 - 2cx^2 - 2bx^2 + 6ax^2 + c^2 - 2bc \\
& - 2ac + b^2 - 2ab + a^2,
\end{align*}\]
and finally
\[\begin{multline}
x^8 - 4\left( a + b + c \right) x^6 + \left( 6c^2 + \left( 4b + 4a \right) c + 6b^2 + 4ab + 6a^2 \right ) x^4 \\
+ 4\left( \left( a + b - c \right) c^2 + \left( b^2 - 10ab + a^2 \right) c - b^3 + ab^2 + a^2b - a^3\right) x^2 \\
+ c^4 - 4\left( a + b \right) c^3 + \left( 6b^2 + 4ab + 6a^2 \right) c^2 + \left( -4b^3 + 4ab^2 + 4a^2b \right. \\
\left. - 4a^3 \right) c + b^4 - 4ab^3 + 6a^2b^2 - 4a^3b + a^4 = 0.
\end{multline}\]
Let's now substitute expressions for $a$, $b$ and $c$ and simplify:
\[\begin{align*}
& 29156245504x^5 - 84494799470592x^4 + 90809255028129792x^3 \\
& - 46242678657209335808x^2 + 11296854647375577219072x \\
& - 1067552764176992047202304 = 0. \end{align*} \]
This quintic can be reduced because we know its one solution, namely $x = 567$ by Opalg . Hence we obtain a quartic:
\[ x^4-2331x^3+1792896x^2-569457920x+64576528128 = 0. \]
The solution of this quartic is a boring algebraic procedure. First one needs to substitute $x = t + 2331/4$ to get rid of the $x^3$ term. Hence the quartic equation can be written as
\[ t^4 - \frac{1957515}{8}t^2 - \frac{504331747}{8}t - \frac{1124218036179}{256} = 0. \]
Next one needs to move $t^1$ and $t^0$ terms to right hand side, and manipulate the left hand side to a square:
\[ \left( t^2 - \frac{1957515}{16} \right) ^2 = \frac{504331747}{8}t + \frac{1239020752851}{64}. \]
To get the right hand side to a square too, one can add an unknown number $q$ to the equation such that left hand side stays a square. In other words: \[ \left( t^2 - \frac{1957515}{16} + q \right)^2 = 2qt^2 + \frac{504331747}{8}t + q^2 - \frac{1957515q}{8} + \frac{1239020752851}{64}. \]
Left hand side is a quadratic function of $t$ now. For this to be a perfect square, one must write its discriminant (with respect to $t$) equal to $0$.
\[ 512q^3 - 125280960q^2 + 9912166022808q - 254350511032072009 = 0. \]
To solve this cubic one first substitutes $q = p + 652505/8$ to remove $q^2$ term. That gives the reduced cubic
\[ p^3 - 597930816p - 2945222125792 = 0. \]
Now, as the cubic has 3 real roots (check out yourself!), one can manipulate the reduced cubic using the formula for cosine, $4\cos ^3 \theta - 3\cos \theta = \cos 3\theta$. To do this, one substitutes $p = u\cos \theta$ and chooses $u$ accordingly:
\[\begin{align}
4\cos ^3 \theta - \frac{2391723264}{u^2}\cos \theta &= \frac{11780888503168}{u^3} \quad \Rightarrow \\
\frac{2391723264}{u^2} &= 3 \quad \Leftrightarrow \\
u &= 10672\sqrt{7} \quad \Rightarrow \\
\cos 3\theta &= \frac{206879}{149408\sqrt{7}} \end{align}\]
Now one can solve $\theta$ and hence finally $q$ by going backwards the substitutions that were made:
\[\begin{align*}
\theta &= \frac{1}{3}\arccos \left( \frac{206879}{149408\sqrt{7}} \right) + \frac{2n\pi}{3} \quad \Rightarrow \\
p &= 10672\sqrt{7} \cos \left( \frac{1}{3}\arccos \left( \frac{206879}{149408\sqrt{7}} \right) + \frac{2n\pi}{3} \right) \quad \Rightarrow \\
q &= \frac{652505}{8} + 10672\sqrt{7} \cos \left( \frac{1}{3}\arccos \left( \frac{206879}{149408\sqrt{7}} \right) + \frac{2n\pi}{3} \right) .
\end{align*} \]
Hence the approximate solutions for the parameter $q$ are:
\[\begin{align*}
q_{n=0} &= q_0 \approx 108182.1818025683 \\
q_{n=1} &= q_1 \approx 60098.83251836192 \\
q_{n=2} &= q_2 \approx 76408.36067906974.
\end{align*}\]
Now, as the number $q$ is known, one can continue the solving of the quartic equation and write its left hand side to a square:
\[ \left( t^2 + q - \frac{1957515}{16} \right) ^2 = 2q\left( t + \frac{504331747}{32q} \right) ^2. \]
After moving terms to right hand side, one can use the formula $a^2 - b^2 = (a+b)(a-b)$:
\[\begin{multline}
\left[ \frac{32q^2 - 3915030q + 504331747\sqrt{2q}}{32q} + \sqrt{2q} \ t + t^2 \right] \\
\cdot \left[ \frac{32q^2 - 3915030q - 504331747\sqrt{2q}}{32q} - \sqrt{2q} \ t + t^2 \right] = 0.
\end{multline} \]
Now these 2 quadratic equations are trivial to solve and one can write down the solutions for the original quartic (and hence the original square root equation) by going backwards the substitutions: \[\begin{align}
x_1 &= \frac{2331}{4} - \frac{ 4\sqrt{2q^3} + \sqrt{2}\sqrt{ -16q^3 + 3915030q^2 - 504331747\sqrt{2q^3}}}{8q} \\
x_2 &= \frac{2331}{4} - \frac{ 4\sqrt{2q^3} - \sqrt{2}\sqrt{ -16q^3 + 3915030q^2 - 504331747\sqrt{2q^3}}}{8q} \\
x_3 &= \frac{2331}{4} + \frac{ 4\sqrt{2q^3} - \sqrt{2}\sqrt{-16q^3 + 3915030q^2 + 504331747\sqrt{2q^3}}}{8q} \\
x_4 &= \frac{2331}{4} + \frac{ 4\sqrt{2q^3} + \sqrt{2}\sqrt{-16q^3 + 3915030q^2 + 504331747\sqrt{2q^3}}}{8q} ,
\end{align}\]
where $q = q_1$ (or $q_0$ or $q_2$, but prefered the smallest absolute value). A closer look to the solutions $x_i$ tells, only $x_4 \approx 1184.131\ldots$ is correct and rest are false solutions due to the squaring of the original equation.

P.S. Posted this from phone... Line breaks can be placed poorly. Sorry about that!