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I solved this problem the fifteen years ago without numerically solving equations of degree 4, I was happy in a substitution that I avoid directly attacking equations of degree 4.

enter image description here Today my nephew, who is an enthusiastic student of mathematics, proposes me the same problem. It occurs to me that I am a very rusted to algebraic substitutions exhaustively. I tried for about 3 hours. How disappointed do not want my nephew to his attempts (and not me) ask for help to the ME community. And of course I will give all the credits to ME.

My attempt. enter image description here

Note that $ \cos \alpha = \frac{1}{x}$, $\cos \alpha = \frac{y}{1}$, $\cos \alpha = \frac{y+1}{x+1}$. Then we have $$ xy=1. $$ By Pythagoras Theorem we have $(y+1)^2+1^2=(x+1)^2 \Longleftrightarrow y^2+2y+1+1=x^2+2x+1$, i.e. $$ x^2-y^2+2(x-y)-1=0. $$ Update [ July 26 2016 ] I remember the time I solved this question, I tried something like \begin{align} x^2-y^2+2(x-y)-1=0 & \Longleftrightarrow (x-y)[x+2+y]=1 \\ & \Longleftrightarrow (x-y)[x+2\cdot 1+y]=1 \\ & \Longleftrightarrow (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})[x+2\sqrt{x}\sqrt{y}+y]=1 \\ & \Longleftrightarrow (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})^3=1 \\ & \Longleftrightarrow (\sqrt{x}+\sqrt{y})^3=\frac{1}{(\sqrt{x}-\sqrt{y})} \\ & \Longleftrightarrow (\sqrt{x}+\sqrt{y})^3=\frac{(\sqrt{x}+\sqrt{y})}{(x-y)} \\ & \Longleftrightarrow (\sqrt{x}+\sqrt{y})^2=\frac{1}{(x-y)} \\ & \Longleftrightarrow (x+y+2)=\frac{1}{(x-y)} \end{align} We then have two ways to tackle the problem: $$ \left\{\begin{array}{rl}\sqrt{x}\sqrt{y}=&1 \\ (\sqrt{x}+\sqrt{y})^3=&\frac{1}{(\sqrt{x}-\sqrt{y})} \end{array}\right. \quad\mbox{ or } \quad \left\{\begin{array}{rl}xy=&1 \\ (x+y+2)=&\frac{1}{x-y} \end{array}\right. $$

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From $y = 1/x,$ then multiplying by $x^2,$ i got $$ x^4 + 2 x^3 - x^2 - 2 x - 1. $$ This looks bad. However, set $$ x = t - \frac{1}{2} $$ and you get rid of the cubic term, always worth a try. I was pleased to discover that the linear term also vanished, giving $$ t^4 - \frac{5}{2} t^2 - \frac{7}{16}, $$ and you can solve for $t^2$ with the Quadratic Formula. I get $$ t^2 = \frac{5 \pm \sqrt {32}}{4} $$ with two pure imaginary roots, a real negative, and a real positive for $t$ itself. Then $x$ is that minus 1/2. I get $1.132241883$ as $x.$

One good habit is to simply draw a graph of the function. I do them by hand with a calculator to find points. I have appended a good online graph. Notice that the graph appears to be symmetric across the vertical line $x = -\frac{1}{2}.$ We could confirm this by taking $ f(x) = x^4 + 2 x^3 - x^2 - 2 x - 1 $ and then checking whether $f(-1-x) = f(x)??$ In turn, this confirmation would tell us that the translation I tried would, in fact, give a graph symmetric across the y axis, meaning all even exponents.

Calculus ideas that are, at least, consistent with the symmetry notion include $$ f'(x) = 2 (2x+1) \left(x^2 + x -1 \right) $$ and $$ f''(x) = 2 \left(6 x^2 + 6 x - 1 \right), $$ so that $x=-1/2$ gives a local maximum, while the inflection points are symmetric around $x = -1/2$ by the quadratic formula.

Huh. Turns out the local minima really are along $y = -2,$ since $$ f(x) + 2 = \left( x^2 + x - 1 \right)^2. $$ Go Figure.

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parisize = 4000000, primelimit = 500509
? x = -1 - w
%1 = -w - 1
? x^4 + 2 * x^3 - x^2 - 2 * x - 1
%2 = w^4 + 2*w^3 - w^2 - 2*w - 1
? 

==================================

parisize = 4000000, primelimit = 500509
? factor( x^4 + 2 * x^3 - x^2 - 2 * x - 1 )
%1 = 
[x^4 + 2*x^3 - x^2 - 2*x - 1 1]

? x = t - (1/2)
%2 = t - 1/2
? p = x^4 + 2 * x^3 - x^2 - 2 * x - 1
%3 = t^4 - 5/2*t^2 - 7/16
? 

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enter image description here

enter image description here

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  • $\begingroup$ +1 For your answer accurate and succinct. However the substitution t = x-0.5, falls from the sky as if by magic. I would appreciate a more constructive response or some intuitive explanation for using such substitution. $\endgroup$ Jul 26 '16 at 12:11
  • $\begingroup$ @MathOverview, it's standard procedure to remove the term of degree $3$ (or $n-1$ in equations of degree $n$). $\endgroup$
    – lhf
    Jul 26 '16 at 12:39
  • $\begingroup$ @lhf Yes this is true. But how to see intuitively that this procedure will also eliminate the term for degree 1? This is the point. $\endgroup$ Jul 26 '16 at 12:48
  • $\begingroup$ @MathOverview, oh, right! $\endgroup$
    – lhf
    Jul 26 '16 at 12:50

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