Solution:

Suppose not. We will obtain a contradiction.

Take any point $A$. Suppose the four edges at $A$ are $BA$, $CA$, $DA$, $EA$. If there is any other point $X$ not joined to any of $A$, $B$, $C$, $D$, $E$ then with $A$ it forms the required pair of points. Suppose the three other points joined to $B$ (apart from $A$) are $B_1$, $B_2$, $B_3$. Similarly $C_i$, $D_i$ and $E_i$. Then all 12 points $B_i$, $C_i$, $D_i$, $E_i$ must be distinct from each other and from $A$, $B$, $C$, $D$, $E$ or there would be a point $X$. Thus, in particular, $A$ is not part of a triangle. But $A$ was arbitrary, so the graph has no triangles. Hence there cannot be an edge $B_iB_j$ (or $C_iC_j$, $D_iD_j$, $E_iE_j$).

We have 4 edges $AX$, 12 edges $BX$, $CX$ etc, and $17 \times 4 / 2 = 34$ edges in all, so there must be 18 edges $B_iC_j$ etc. Each gives a different cycle length 5 through $A$ (e.g. $ABB_iC_jC$). The same argument shows that every point must lie on 18 cycles length 5. Hence there must be a total of $17 \times 18 / 5$ such cycles. Contradiction.