Igor wrote: Do there existe $5$ points in the space such that for all $n\in\{1,2,\ldots,10\}$ there exist two points such that the distance between them is $n$? 5 points $\Rightarrow $ exactly 10 distances point to point $\Rightarrow $ each ten distances 1 to 10 occur just once. Then let $XY=1$ and Z another point. WLOG say $XZ>YZ$ (equality is impossible, as said above). Then, in triangle XYZ, $XZ\leq XY+YZ$ $\Rightarrow $ $0<XZ-YZ\leq XY=1$ and, since XZ and YZ are integers, $XZ=YZ+1$ $\Rightarrow $ X,Y and Z are colinear $\Rightarrow $ All five points, if they exist, are colinear. Then, WLOG say the five colinear points are, in this order, A,B,C,D and E, with $AE=10$. The sum of the ten distances is : $S=AB+(AB+BC)+(AB+BC+CD)+10+BC+(BC+CD)$ $+(10-AB)+CD+(10-AB-BC)+(10-AB-BC-CD)$ $S=40+2BC+2CD$, even, which can't be equal to $1+2+3+4+5+6+7+8+9+10=55$, odd. So such 5 points don't exist. -- Patrick

pco wrote: Igor wrote: Do there existe $5$ points in the space such that for all $n\in\{1,2,\ldots,10\}$ there exist two points such that the distance between them is $n$? 5 points $\Rightarrow $ exactly 10 distances point to point $\Rightarrow $ each ten distances 1 to 10 occur just once. Then let $XY=1$ and Z another point. WLOG say $XZ>YZ$ (equality is impossible, as said above). Then, in triangle XYZ, $XZ\leq XY+YZ$ $\Rightarrow $ $0<XZ-YZ\leq XY=1$ and, since XZ and YZ are integers, $XZ=YZ+1$ $\Rightarrow $ X,Y and Z are colinear $\Rightarrow $ All five points, if they exist, are colinear. Then, WLOG say the five colinear points are, in this order, A,B,C,D and E, with $AE=10$. The sum of the ten distances is : $S=AB+(AB+BC)+(AB+BC+CD)+10+BC+(BC+CD)$ $+(10-AB)+CD+(10-AB-BC)+(10-AB-BC-CD)$ $S=40+2BC+2CD$, even, which can't be equal to $1+2+3+4+5+6+7+8+9+10=55$, odd. So such 5 points don't exist. -- Patrick My way for end: $9$ can only be formed if $1$ is outside, also $2$ outside then or we can't form $8$ and then we have 2 cases with no solutions.

If possible, let the Pentagon formed be non-degenerate. Since $\binom{5}{2}=10$, every pair is connected. But, by triangle inequality, if one of the lengths is $1$ and the other sides of triangle are integers, then not possible. So some 3 points are collinear and dist between 2 of them is $1$. Let the points be $X,Y,Z$ such that $Y$ is between $X,Z$ and $XY=1$. Now let the other 2 points be $P,Q$ not on line $XYZ$. Again $PXY$, $QXY$ don't satisfy triangle inequality so all points are collinear, say in the order $PQXYZ$. Now $PQ \neq QX \neq XY \neq YX$ so their sum is at least $1+2+3+4+5=15>10$ not possible so done.