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I came across this olympiad algebra problem, asking to solve for $x$:

$x\ +\ \sqrt { (x+1)(x+2) } \ +\ \sqrt { (x+2)(x+3) } +\ \sqrt { (x+3)(x+1) } =\ 4$

Here was my try:

If $$x\ +\ \sqrt { (x+1)(x+2) } \ +\ \sqrt { (x+2)(x+3) } +\ \sqrt { (x+3)(x+1) } =\ 4$$

Then $\quad \sqrt { (x+1)(x+2) } +\sqrt { (x+2)(x+3) } +\sqrt { (x+3)(x+1) } =4-x$.

Further, I tried squaring the equation on both sides, but that doesn't seem to solve my problem. Please help.

Thank you.

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    $\begingroup$ The octic method described in my answer can be used to show that for any rational constant $a,b,c$, then $$(x+a)(x+b) = y_1^2\\(x+a)(x+c) = y_2^2\\(x+b)(x+c) = y_3^2$$ has an infinite number of rational solutions $x$. $\endgroup$
    – Tito Piezas III
    Dec 20, 2015 at 17:55

2 Answers 2

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Hint: Observe that $:x = (x+1) + (x+2) - (x+3)$. This leads us to letting $a = \sqrt{x+1}, b = \sqrt{x+2}, c = \sqrt{x+3} \Rightarrow a^2+b^2-c^2 + ab+bc+ca = 4, b^2-a^2 = 1 = c^2-b^2 \Rightarrow a^2-1 +ab+bc+ca= 4 \Rightarrow (a+b)(a+c) = 5 \Rightarrow \dfrac{a+c}{b-a} = 5 \Rightarrow a+c = 5b-5a\Rightarrow 6a = 5b-c\Rightarrow 6\sqrt{x+1} = 5\sqrt{x+2} - \sqrt{x+3}$. Can you continue?

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    $\begingroup$ There is a second solution. Kindly see graph in my answer. $\endgroup$
    – Tito Piezas III
    Dec 20, 2015 at 7:28
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While Kf-Sansoo has given an elegant answer, if the problem asks for any real (and not just rational) solution, then it misses a second one which is a root of a quartic equation, hence normally is not easy to do by hand.

In general, the two solutions to,

$$x+\sqrt{(x+1)(x+2)}+\sqrt{(x+2)(x+3)}+\sqrt{(x+3)(x+1)} = n\tag1$$

for real $n>0$ are,

$$x = \frac{(n^2+4n+5)^2}{4(n+1)(n+2)(n+3)}-2\tag{2a}$$

and the appropriate root of,

$$-23 + 48 n - 22 n^2 + n^4 - 4 (30 - 33 n + 6 n^2 + n^3) x \\+ 16 (-11 + 6 n) x^2 + 16 (-6 + n) x^3 - 16 x^4=0\tag{2b}$$

For $n = 4$, we have $x_1 = -311/840 \approx -0.37$. Then $x_2 \approx -5.12357$ as a root of,

$$73 - 232 x + 208 x^2 - 32 x^3 - 16 x^4 = 0$$

with both valid for the positive case of $\sqrt{z}$ as the graph from Walpha below shows,

enter image description here

$\color{green}{Edit:}$ When using Kf-Sansoo's method, we end up with an expression of form,

$$\prod^4 (c_1\sqrt{x+1}\pm c_2\sqrt{x+2}\pm c_3\sqrt{x+3}) = 0$$

Let $n=4$ and we get $x = -311/840$. Simpler, but the price to pay is we lose a second solution. Another method is to form an octic,

$$\prod^8 \big(y-(\pm\sqrt{z_1}\pm \sqrt{z_2}\pm \sqrt{z_3})\big)=0 \tag3$$

After it is formed, substitute into $(3)$ the ff,

$$y = n-x\\z_1=(x+1)(x+2)\\z_2=(x+2)(x+3)\\z_3=(x+3)(x+1)$$

and we get linear/quartic factors given by $(2a), (2b)$. Less simpler, but it yields a second valid solution.

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  • $\begingroup$ Just by curiosity, how did you arrive to thse general expressions ? Are they valid for any $n$ (real or integer only ?). Thanks $\endgroup$
    – Claude Leibovici
    Dec 20, 2015 at 6:33
  • $\begingroup$ @ClaudeLeibovici: I used the octic formed by $y-(\pm\sqrt{z_1}\pm\sqrt{z_2}\pm\sqrt{z_3})$, hence to completely solve the problem by hand is too much. The two solutions should be valid for any real $n$, provided one uses the correct signs of $(1)$ and the appropriate root of $(2b)$. $\endgroup$
    – Tito Piezas III
    Dec 20, 2015 at 6:41
  • $\begingroup$ Thank you ! It is impressive. I think that it could be good you add these explanations in your answer. Cheers. $\endgroup$
    – Claude Leibovici
    Dec 20, 2015 at 6:47
  • $\begingroup$ @ClaudeLeibovici: And I've added a graph which shows the two solutions. $\endgroup$
    – Tito Piezas III
    Dec 20, 2015 at 7:27
  • $\begingroup$ Thank you ! You improved a lot !! $\endgroup$
    – Claude Leibovici
    Dec 20, 2015 at 7:30

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