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Find the unique solution to

$$ 5\,ninja + retook + quarter + turn = \_\,\_\,\_\,\_ + \_\,\_\,\_\,\_\,\_\,\_\,\_\,\_\,\_\,\_ $$

where each unknown is a different _ _ _ _  _ _ _ _ _ _ _ _ _ _. Granted, way more than eight _ _ _ _  _ _ _ _ _ _ _ _ _ _ s exist, but some are special.

P. S. In case there are technical problems displaying the underscores, the hidden words consist of 4 and 10 letters respectively.

P. P. S. You must be very precise in answering exactly the question that is posed. Though the most important thing of course is your narration. k?

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  • $\begingroup$ Does this have something to do with quaternions? $\endgroup$ – user574848 Jan 22 at 8:14
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    $\begingroup$ @user574848: 👍 $\endgroup$ – Roman Odaisky Jan 23 at 13:26
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    $\begingroup$ Thought so, though I don't know much about them apart from that they exist so I probably won't be able to solve this $\endgroup$ – user574848 Jan 23 at 19:23
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Partial answer (I'm missing something at the end).

The first thing to notice is that

There are $11$ variables and apparently only one equation (in fact, five, as explained below, but that is still too few for a complete solution), but "You must be very precise in answering exactly the question that is posed." The question that is posed is at the end: "k?" We're only looking for one variable. That's a relief.

The _ _ _ _ _ _ _ _ _ _ _ _ _ _ stand for

UNIT QUATERNION

Wait what is that?

A quaternion is a kind of number that generalizes imaginary numbers such as $2+3i$, involving two more symbols $j$ and $k$, all subject to the defining relations $$i^2=j^2=k^2=ijk=-1$$ which Hamilton famously carved into the stone of a Dublin bridge where he was standing when he discovered it. An arbitrary quaternion looks like $$a+bi+cj+dk$$ where $a, b, c, d$ are real numbers.

And the first word?

A unit quaternion now, is a quaternion $a+bi+cj+dk$ whose squared norm $a^2+b^2+c^2+d^2$ is equal to $1$. To go further, we will need two useful properties of units: $$u \text{ is a unit } \Rightarrow -u \text{ is a unit } $$ $$u, v \text{ are units } \Rightarrow uv \text{ is a unit } $$

The last thing we need to know before solving the equation:

The triangle inequality holds for quaternions. This implies that if five quaternions $A, B, C, D, E$ are units and their sum has norm $5$ (the largest possible norm a sum of five units can have), then all of these quaternions are equal to each other. (Roughly, they must point to the same direction, and since they have the same magnitude they must be equal.) Moreover, if $A+B+C+D+E=5F$ where $F$ is a unit, then $5A=5F$ and we get that $F$ is also equal to $A$ (and $B$, etc.).$$$$ This property, together with the $5$ stuff = a sum of five things having a unique solution, is what points at some kind of Euclidean geometry. The "eight" from the hint then points at unit quaternions, because of the eight elementary units: $1, i, j, k$ and their negatives.

Finally solving the equation:

$$5 ninja = unit + quaternion - retook - quarter - turn$$ where every word is a product of units and therefore a unit itself, implies that $$ninja =unit =quaternion = -retook =-quarter=-turn$$ From there one can derive the value of $k$ in four ways (one for each of the four other words). For instance $$ \begin{align}&&retook&=turn\\&&k&=(retoo)^{-1}turn &&\text{"retoo inverse turn"}\\\text{or}&&k&=(retoo)^*turn&&\text{"retoo star turn"} \end{align}$$ since the inverse of a unit is also equal to its conjugate. Note that $(retoo)^*$ can be rewritten $o^*o^*t^*e^*r^*$. There is maybe a fancy way to read that (?).

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  • $\begingroup$ Nice! However, these sorts of simplifications are unfortunately not possible because multiplication is not commutative. $\endgroup$ – Roman Odaisky Feb 12 at 23:43
  • $\begingroup$ @RomanOdaisky Of course, stupid me. $\endgroup$ – Arnaud Mortier Feb 13 at 7:45
  • $\begingroup$ @RomanOdaisky Fixed. Still missing something, I think. $\endgroup$ – Arnaud Mortier Feb 13 at 8:47
  • $\begingroup$ I wonder if the $i$,$j$, and $k$ are the special units rather than variables? Then, perhaps, some simplification is possible $\endgroup$ – Dr Xorile Feb 25 at 0:23
  • $\begingroup$ @DrXorile I thought about that too, but then the question that is asked is trivial. Then again this might be the point. $\endgroup$ – Arnaud Mortier Feb 25 at 8:38

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