Help with my proof of maximum number of intersection points of n lines being n(n-1)/2 I know that this question is a hoary old chestnut, but I have never seen a proof before working one out myself, so I'd like you to help me see if mine is rigorous enough.  

Obviously with $n$ linear two-variable equations, a pair has at most one unique solution, so the maximum cannot exceed the number of pairs that can be chosen is $\frac{N(N-1)}{2}$

Then one proceeds to construct a system in which each pair has a unique solution distinct from the others. 

Let $a_k(x)+b_k=y$ for each $k\in [1, N]$. Let $a_k$ be $1/p_k$, the reciprocal of the $k$-th prime. Let $b_k$ be $1/p_k^2$, i.e. the square of the reciprocal of the $k$-th prime.  

For the pair $k=$ distinct $i$ and $j$, the solution is $x=-(1/p_i+1/p_j)$. (A bit of algebra will figure that out.) The proof consists of showing that the sum of the reciprocal of two distinct primes cannot be represented by such a sum with two other primes.

$1/p_i+1/p_j=(p_i+p_j)/(p_i*p_j)$, the RHS in its simplest form already since if $GCD(a,b)=1$, $GCD(a+b,ab)=1$ too. (Otherwise some prime divides $ab\Rightarrow$ some prime divides one of $a$ or $b\Rightarrow$ that prime divides the other of $a$ or $b$ $\#$). 

Thus for any $p_m, p_n$ that satisfies $1/p_m+1/p_n=(p_i+p_j)/(p_i*p_j)$, it follows that $p_m+p_n=p_i+p_j$, and that $(p_m)(p_n)=(p_i)(p_j)$. If $p_i$ and $p_j$ are known, then the two above equations imply $p_m$ and $p_n$ equal $p_i$ and $p_j$, except in respect of the order. (To be really rigorous on this point, construct a quadratic equation the roots of which are $p_m$ and $p_n$).

Thus for two distinct pairs of equations, the solution cannot overlap.

Thanks for any help on this matter!
So if you pick $n$ lines at random they will intersect in $\binom{n}{2}$ points with probability 1, so there are all sorts of ways you could construct such an example.  I didn't read your exact construction, but here's a simple way:

The result is trivial for $n=1$.  Suppose we have a set of n lines, each of which intersects each of the others in a unique point.  Now:


Pick a slope $m$ which does not match that of any of the other n lines.
Pick a line $\ell$ of slope $m$ which does not pass through any of the $\binom{n}{2}$ points of intersection between the other lines.
Add $\ell$ to our set of lines.


Then our new set consists of $n+1$ lines, with all pairwise intersections unique.
So if you pick $n$ lines at random they will intersect in $\binom{n}{2}$ points with probability 1, so there are all sorts of ways you could construct such an example.  I didn't read your exact construction, but here's a simple way:

The result is trivial for $n=1$.  Suppose we have a set of n lines, each of which intersects each of the others in a unique point.  Now:


Pick a slope $m$ which does not match that of any of the other n lines.
Pick a line $\ell$ of slope $m$ which does not pass through any of the $\binom{n}{2}$ points of intersection between the other lines.
Add $\ell$ to our set of lines.


Then our new set consists of $n+1$ lines, with all pairwise intersections unique.