Sums and differences of distinct factors

Question

Given $k, n \in \mathbb{N}$, let $\tau_{k}(n)$ denote the $k$th positive factor of $n$ in strictly increasing order.

For example, $\tau_{1}(6) = 1; \tau_{2}(6) = 2; \tau_{3}(6) = 3; \tau_{4}(6) = 6$.

Question. For what $n$ are there distinct $a, b, p, q$ such that:

1. $\tau_{a}(n) + \tau_{b}(n) = \tau_{p}(n) - \tau_{q}(n)$, and

2. $\tau_{a}(n)\tau_{b}(n) = n = \tau_{p}(n)\tau_{q}(n)$

For example, $n = 6$ works with $a = 2; b = 3; p = 4; q = 1$:

This is because $2 + 3 = 6 - 1$, and $2 \cdot 3 = 6 = 6 \cdot 1$.

Notes:

• $\tau_1(n)=1$ for all $n$
• If $a,b,p,q$ exist for $n$ then $a',b',p',q'$ exist for $n\cdot m^2$ for all $m \in \mathbb{Z}$ and $a'=am; b'=bm; p'=pm; q'=qm$
• Computing a few by hand, the admissible $n$ seem to sync up with OEIS A009112

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More generally: For all $n\in \mathbb{N}$, we can assign a value $$\hat{\tau}(n)=\text{min}[\tau_{a}(n) + \tau_{b}(n)] - [\tau_{p}(n) - \tau_{q}(n)]$$

for distinct $a, b, p, q$ subject to the constraint $\tau_{a}(n)\tau_{b}(n) = n = \tau_{p}(n)\tau_{q}(n)$, where $\hat{\tau}=0$ for all values that satisfy the constraints mentioned above.

Does calculating $\hat{\tau}(n)$ or looking at the sequence of $\hat{\tau}(n)$s tell us anything meaningful?

Answer 1

Since the indices of the factors in the strictly increasing order don't actually play any role, I'll use $A,B,P,Q$ to denote the actual factors instead of $\tau_a(n)$, $\tau_b(n)$, $\tau_p(n)$ and $\tau_q(n)$. The condition given can then be rewritten as $AB=PQ=n$ and $A+B=P-Q$.

Let's start by showing that all numbers from OEIS A009112 admit such factorizations. The terms of the sequence are areas of Pythagorean triangles and it is known that all such triangles have sides expressible as $$\begin{eqnarray} k & \cdot & (u^2-v^2) \\ k & \cdot & (2uv) \\ k & \cdot & (u^2+v^2) \end{eqnarray}$$ for some positive integers $k$ and $u>v$, with $\gcd(u,v)=1$. Now, consider $$\begin{array}{cc} A=k\cdot v\cdot (u+v) & P=k\cdot u\cdot (u+v) \\ B=k\cdot u\cdot (u-v) & Q=k\cdot v\cdot (u-v) \\ \end{array}$$ and see that these factors satisfy the given conditions and the products $AB$ and $PQ$ are both equal to the area of the corresponding triangle.

In the other direction, let $A,B,P,Q$ satisfying the conditions be given to us. Then, the integers $(A+Q)$, $(B+Q)$, and $(A+B)$ form a Pythagorean triple. Let's check it:

$$\begin{eqnarray}(A+Q)^2 + (B+Q)^2 & = & A^2 + B^2 + 2(A+B)Q + 2Q^2\\ & = & A^2 + B^2 + 2(P-Q)Q + 2Q^2 \\ & = & A^2 + B^2 + 2PQ \\ & = & A^2 + B^2 + 2AB \\ & = & (A+B)^2\\ \end{eqnarray} $$

The area of this triangle is $$\begin{eqnarray} \frac{1}{2}(A+Q)(B+Q) & = & \frac{1}{2}(AB + (A+B)Q + Q^2) \\ & = & \frac{1}{v2}(AB + (P-Q)Q + Q^2) \\ & = & \frac{1}{2}(AB + PQ) \\ & = & AB = PQ = n\\ \end{eqnarray}$$

Thus, the numbers $n$ admitting two distinct factorizations satisfying the given conditions are exactly those in sequence A009112.