Alexandrian Integers

We shall call a positive integer $A$ an “Alexandrian integer”, if there exist integers $p, q, r$ such that:
$$A = p \cdot q \cdot r$$
$$\dfrac{1}{A} = \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r}.$$
For example, $630$ is an Alexandrian integer ($p = 5, q = -7, r = -18$).
In fact, $630$ is the $6$th Alexandrian integer, the first $6$ Alexandrian integers being: $6, 42, 120, 156, 420$, and $630$.
Find the $150000$th Alexandrian integer.

This problem requires knowledge of solving Diophantine equations and number theory.

Let’s start by rewriting the Alexandrian integer equation. We simplify the equation $\dfrac{1}{A} = \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{r}$ and get, $pqr = p(q + r) + qr$. This can be rearranged into the diophantine equation $A = pqr = (q+r)p + qr$.

Let’s make two assumptions:
1. $p \leq q \leq r$ (since $p$, $q$, and $r$ are interchangeable)
2. $p > 0$ (since values can be negative)

The assignment works for the six first Alexandrian numbers.

If one assumes $r$ to be $p + k$ with $k \geq 1$ (since by 1. $r \geq q \geq p$), one can follow with analogous computations to yours:

$A = pqr = p(p+q)(p+k) = p^2(q+k) + p^2k = 2p^3 + (q+k-2)p^2 + p^2k$. This equation has to be solved or inspected for $k$, after fixing $p$.

Now here comes the most important insight: once one has guessed or written down the equations for $p = 1, 2, 3, …$, one can see a pattern for the minimal $k$: If $p$ is odd, then $k_min = 2$. If $p$ is even, $k_min = 1$. For these selections of $k$, $q == p$.

So, let’s find how many numbers for a single $p$ can be generated, if $k_{min} \leq k \leq 2p$ holds: For odd $p$, there are exactly $p$ different k, for even $p$ there are $p+1$ different values of k. In total there are $2* \sum_{k=1}^{n/2} k + \sum_{k=1}^{n/2} k = \sum_{k=1}^{n} k = \dfrac{n*(n+1)}{2}$.

We now solve $\dfrac{n*(n+1)}{2} > 150000$ for $n$, using that $n^2$ for $n > 270$ (take the square root of both sides of the inequality). Though one can directly compute the exact solution $n = 549$, it is faster to guess that $n = 550$ to solve $n*(n+1) = 300000$. Since every $p$ gives at least $p$ solutions, and we only need two solutions from $p = 550$, we don’t even have to compute the actual number of solutions arising from $p = 549$.

This gives us the Alexandrian integer number $150000$ is given by $p = 550$ and $k = 2$, hence
$A = p*p*(p+2)^2 = 550^2*(550+2)^2 = 166926000400$ is the $150000$th Alexandrian number.

Therefore, the answer to the problem is $166926000400$.

More Answers:
Perfect Right-angled Triangles
Skew-cost Coding
Heighway Dragon

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