Counting Solutions of 1/m ± 1/n = 1/k (TNF Trick)

📘 One-line factorisation

\[ \frac1m + \frac1n = \frac1k \quad (k\in\mathbb Z_{>0}) \ \Longleftrightarrow\ (m-k)(n-k)=k^2 \] \[ \frac1m - \frac1n = \frac1k \quad (k\in\mathbb Z_{>0}) \ \Longleftrightarrow\ (m-k)(n+k)=-\,k^2 \]

Here \(\mathrm{TNF}(N)\) means “total number of factors of \(N\)” (ordered divisor-pair counting uses these factors naturally).

🧮 Counting formulas (TNF trick)

Total integral solutions (both cases): \(\;2\cdot \mathrm{TNF}(k^2)-1\).
For \(\frac1m+\frac1n=\frac1k\):
- Positive integral solutions \(=\ \mathrm{TNF}(k^2)\).
- Negative integral solutions \(=\ 0\).
- “One + & one −” \(=\) Total − Positive.
For \(\frac1m-\frac1n=\frac1k\):
- Positive integral solutions \(=\ \dfrac{\mathrm{TNF}(k^2)-1}{2}\).
- Negative integral solutions \(=\ \dfrac{\mathrm{TNF}(k^2)-1}{2}\).
- “One + & one −” \(=\) Total − \(2\times\)Positive.

If \(k=\prod p_i^{a_i}\), then \(\ \mathrm{TNF}(k^2)=\prod (2a_i+1)\).

🧠 Why TNF\((k^2)\) appears (proof sketch)

Plus case

\((m+n)/mn=1/k \Rightarrow mk+nk=mn \Rightarrow (m-k)(n-k)=k^2\). For each ordered factor pair \(d\cdot e=k^2\), set \(m=d+k,\ n=e+k\). Positive pairs \((d,e>0)\) give all-positive solutions ⇒ count \(=\mathrm{TNF}(k^2)\). Including negative factors yields total \(=2\mathrm{TNF}(k^2)-1\) (central pair double-count removed).

Minus case

\((n-m)/mn=1/k \Rightarrow kn-km=mn \Rightarrow (m-k)(n+k)=-k^2\). Take \(d\cdot e=-k^2\) and set \(m=d+k,\ n=e-k\). The negative product splits evenly ⇒ positive solutions \(=(\mathrm{TNF}-1)/2\), same for negative.

✅ Worked examples

\(\frac1m+\frac1n=\frac1{12}\): \(12=2^2\cdot3\Rightarrow \mathrm{TNF}(12^2)=(5)(3)=15\). Positive \(=15\), Total \(=2\cdot15-1=29\), One \(+\)/One \(−\) \(=29-15=14\).
\(\frac1m-\frac1n=\frac1{20}\): \(20=2^2\cdot5\Rightarrow \mathrm{TNF}(20^2)=(5)(3)=15\). Positive \(=(15-1)/2=7\), Negative \(=7\), Total \(=29\), One \(+\)/One \(−\) \(=29-14=15\).

🗂️ 60-second exam playbook

Prime-factorise \(k\). Compute \(\mathrm{TNF}(k^2)=\prod(2a_i+1)\).
Pick case:
- Plus: Total \(=2\mathrm{TNF}-1\), Positive \(=\mathrm{TNF}\), Negative \(=0\), Mixed \(=\) Total − Positive.
- Minus: Positive \(=(\mathrm{TNF}-1)/2\), Negative \(=(\mathrm{TNF}-1)/2\), Total \(=2\mathrm{TNF}-1\), Mixed \(=\) Total − \(2\cdot\)Positive.
To construct an actual pair, use the factorisations above.

⚠️ Edge cases & notes

\(k>0\) integer is assumed. \(k=0\) is invalid for integer \(m,n\).
When the question restricts \(m,n>0\), use the “Positive” counts only.
Counts are for ordered pairs \((m,n)\) (swap usually gives a new solution unless \(m=n\)).