Question 6 of the 2nd round of the 15th Iran mathematical olympiad.

We first show that not every point can be reached from $(0,0)$ for $n le 4$. That it can be done for $n ge 5$ will be shown afterwards. That this solves the general problem follows from the symmetry of the "move from A to B" definition.

Part 1: $n le 4$

Let the rational units be defined as

$$RU_n={r=(r_1,ldots,r_n) in P| text{ satisfying } |r|^2=r_1^2+ldots+r_n^2=1}.$$

A move can be made from $Ain P$ to $B in P$ iff $B-A in RU_n$, where the difference is made component-wise. That means the points in $P$ that can be reached from $(0,0)$ in a finite series of moves are exactly those that can be expressed as a finite sum

$$x=sum_{i=1}^k x_i, text{ where } forall i in {1,ldots,k}: x_i in RU_n. tag{*}label{eq0}$$

We'll now show no component of a rational unit can have a denominator divisible by $4$ (in reduced form) if $n le 4$.

Assume otherwise, then there exists a $r in RU_n$ and writing $forall i=1,ldots,n: r_i=frac{a_i}{b_i}$ in reduced form, we assume w.l.o.g that $b_1$ contains (one of) the highest powers of $2$ among the $b_i$, so $b_1=2^kb'_1, 2nmid a_1, 2 nmid b'_1, k ge2 $. Note that $d=lcm (b_1,ldots,b_n)$ is then of the form $d=2^ko$ with odd $o$. Using the definition of rational units we get

$$sum_{i=1}^n frac{a_i^2}{b_i^2}=1$$

and if we multiply this with $d^2$ we get

$$ sum_{i=1}^n c_i^2=d^2, text { where } c_i=a_ifrac{d}{b_i}in mathbb Z. tag{1} label{eq1}$$

As we noted above $2nmid a_1$ and since $frac{d}{b_1}=frac{o}{b'_1}$ is also odd, so is $c_1$! Since $n le 4$, we can add $0^2$ terms to the left side of eqref{eq1} to fill up to 4 summands if necessary, sowie finally get

$$sum_{i=1}^4 c_i^2 = d^2=16d'^2 tag{2} label{eq2}$$

Considering eqref{eq2} mod 4, and remembering that $0$ and $1$ are the only possible quadratic residues mod 4, the only solutions are $c_1^2equiv c_2^2 equiv c_3^2 equiv c_4^2 equiv 0 pmod 4$ and $c_1^2equiv c_2^2 equiv c_3^2 equiv c_4^2 equiv 1 pmod 4$. Since we know that $c_1$ is odd, only the second option is possible, so all the $c_i$ are odd.

Considering eqref{eq2} now mod 16, and remembering that the only odd quadratic residues mod 16 are $1$ and $9$, we see that eqref{eq2} is impossible to fullfil with all odd $c_i$ mod 16, because the only options are $4times 1, 3times 1 + 1times 9, 2times 1 + 2times 9,1times 1 + 3times 9$ and $4times 9$, which are $equiv 4, 12, 4, 12, 4 pmod {16}$ in that order, so never $0$.

Ok, so we proved that no component of a rational unit has a denominator that is a multiple of $4$. But eqref{eq0} shows that each component of a point that is reachable from $(0,0)$ is the sum of components of rational units. But such a sum can never be $frac14$. Because if it were, then

$$frac14=sum_{i=1}^k frac{a_i}{b_i},$$

where $frac{a_i}{b_i}$ is the reduced form of a component of a rational unit, so $4nmid b_i$ and hence $4nmid lcm(b_1,ldots,b_k)$. Multiplying the above equation with $lcm(b_1,ldots,b_k)$ gives an integer result on the right hand side, but not an integer on the left hand side, as $lcm(b_1,ldots,b_k)$ cannot "cancel" the $4$ in the denominator.

So any point in $P$ where one cordinate is $frac14$ cannot be reached from $(0,0)$ for $n le 4$.

This concludes the proof for part 1.

Part 2: $n ge 5$

We use an old theorem from number theory, Lagrange's four-square theorem, that every natural number can be represented as the sum of four integer squares.

That means for every integer $k ge 1$ we have for integers $a_k, b_k, c_k, d_k$ such that

$$a_k^2 + b_k^2+ c_k^2 + d_k^2 = 4k^2 -1,$$

and thus

$$left(frac{1}{2k}right)^2 + left(frac{a_k}{2k}right)^2 + left(frac{b_k}{2k}right)^2 + left(frac{c_k}{2k}right)^2 + left(frac{d_k}{2k}right)^2 = 1.$$

That means for $nge 5$ there exists a rational unit $r_1$ with $frac1{2k}$ as first component (if $n > 5$, just fill up with zeros). Since the definition of the rational unit only uses the square of each component, changing any number of signs in a rational unit makes the resulting number also a rational unit. Let $r_2$ be the rational unit obtained from $r_1$ by flipping all the signs, expcept on the first component. That means

$$r_1+r_2=(frac1k,0,0,ldots).$$

Obviously $((-r_1) + (-r_2)=(-frac1k,0,0,ldots)$, and similar by permutating the components rational units can be found such that the sum of 2 of them produces the zero vector except for a given component, where the component is $frac1k$ or $-frac1k$.

But each vector from $pin P$ can be obtained as a finite sum of vector with only one component.

$$p=(p_1,ldots,p_n) = (p_1,0,ldots,0) + (0,p_2,0,ldots) + ldots + (0,ldots,0,p_n), tag{3}label{eq3}$$

and if $p_1=frac{a}{b}, b > 0$, then

$$(p_1,0,ldots,0)=sum_{i=1}^{|a|}(pmfrac1b,0,ldots,0),$$

with the signs before the fraction chosen the same as $a$. The same can of course be done for the other summands in eqref{eq3}. But the summands thus calculated are exactly of the type that we proved are the sum of 2 rational units, so each $p=(p_1,ldots,p_n) in P$ is the sum of a finite number of rational units, so can be reached from $(0,0)$.

This concludes part 2 of the proof.