Can You Crack This Week's Rational Equation Challenge?

[anemone]

Here is this week's POTW:

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Solve for the rational solution of the equation \(\displaystyle x+\sqrt{(x+1)(x+2)}+\sqrt{(x+2)(x+3)}+\sqrt{(x+3)(x+1)}=4\).-----

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[anemone]

Congratulations to Opalg for his correct solution(Cool), which you can find below:

Spoiler

Let $f(x) = x + \sqrt{(x+2)(x+3)} + \sqrt{(x+3)(x+1)} + \sqrt{(x+1)(x+2)}$. We are looking for a rational number $x$ such that $f(x) = 4$.

Let $x = \frac cd$, where $c$ and $d$ are integers. Then $$f\bigl(\frac cd\bigr) = \frac{c + \sqrt{(c+2d)(c+3d)} + \sqrt{(c+3d)(c+d)} + \sqrt{(c+d)(c+2d)}}d.$$ For that expression to be rational, it looks as though each of the integers $c+d$, $c+2d$, $c+3d$ will have to be a square. So we are looking for three squares that form an arithmetic progression with first term $c+d$ and common difference $d$. The smallest such set consists of the numbers $1$, $25$, $49$, which form an arithmetic progression with common difference $24$.

So suppose we let $d = 24$ and $c+d = 1$. Then $c = -23$, and \(\displaystyle f\bigl(\frac cd\bigr) = \frac{-23 + 35 + 7 + 5}{24} = \frac{24}{24} = 1.\) Unfortunately that is not equal to $4$. But it looks like a sufficiently promising approach to be worth investigating further. An internet search for "squares in arithmetic progression" reveals that there is a sequence of triples $(u,v,w)$ with the property that $u^2$, $v^2$, $w^2$ form an arithmetic progression with common difference $d$. The first few triples are $$\begin{array}{|c|c|c}u,v,w & u^2,v^2,w^2 & d \\ \hline 1,5,7 & 1,25,49 & 24 \\ 7,13,17 & 49,169,289 & 120 \\ 7,17,23 & 49,289,529 & 240 \\ 17,25,31 & 289, 625, 961 & 336 \\ 31,41,49 & 961,1681,2401 & 720 \\ 23,37,47 & 529,1369,2209 & 840. \end{array}$$ For each of these triples, let $c = u^2-d$, $x = \frac cd$, and compute $f\bigl(\frac cd\bigr)$ in exactly the same way as for $f\bigl(\frac{-23}{24}\bigr)$: $$\begin{array}{|c|c|c|c}u,v,w & d & c & f\bigl(\frac cd\bigr) \\ \hline 1,5,7 & 24 & -23 & \frac{-23 + 35 + 7 + 5}{24} = \frac{24}{24} = 1 \\ 7,13,17 & 120 & -71 & \frac{-71 + 221 + 91 + 119}{120} = \frac{360}{120} = 3 \\ 7,17,23 & 240 & -191 & \frac{-191 + 391 + 119 + 161}{240} = \frac{480}{240} = 2 \\ 17,25,31 & 336 & -47 & \frac{-47 + 775 + 425 + 527}{336} = \frac{1680}{336} = 5 \\ 31,41,49 & 720 & 241 & \frac{241 + 2009 + 1271 + 1519}{720} = \frac{5040}{720} = 7 \\ 23,37,47 & 840 & -311 & \frac{-311 + 1739 + 851 + 1081}{840} = \frac{3360}{840} = 4. \end{array}$$ So if $\boxed{x = -\dfrac{311}{840}}$ then $f(x) = 4$.

I do not know whether that solution is unique, as the question seems to imply.

[This is an appendix to my previous solution.]

Edit. I have since found that the triple $(1,29,41)$ has the property that $1^2,29^2,41^2$ is an arithmetic progression of squares with common difference $d = 840$, and $c = -839$. That gives $x = -\frac{839}{840}$, and $f(x) = \frac{-839 + 1189 + 29 + 41}{840} = \frac{420}{840} = \frac12.$ So my conjecture was wrong, and $f(x)$ is not always an integer.

Alternate solution from other:

Spoiler

Observe that

\(\displaystyle x=(x+1)+(x+2)−(x+3)\)

So we let

\(\displaystyle a=\sqrt{x+1}\\b=\sqrt{x+2}\\c=\sqrt{x+3}\\\)

and upon substituing these functions to the given equality we have

\(\displaystyle a^2+b^2−c^2+ab+bc+ca=4\)

and

\(\displaystyle b^2−a^2=1=c^2−b^2\implies (a+b)(b-a)=1\\a^2−1+ab+bc+ca=4\\(a+b)(a+c)=5\\ \dfrac{a+c}{b-a}=5\\ a+c=5b-5a\\ 6a=5b-c\\6\sqrt{x+1}=5\sqrt{x+2}-\sqrt{x+3}\)

Solving this for $x$ we get $x=-\dfrac{311}{840}$.