Factors of Integer Congruent to 5 modulo 6

Theorem

Let $m$ be an positive integer.

Let $m \equiv 5 \pmod 6$.

Then $m$ has two divisors whose sum is divisible by $6$.

Proof

The validity of the material on this page is questionable. In particular: This is so trivial it's pointless. Both $m$ and $1$ are factors, totalling $6 n$, trivially a multiple of $6$. I expect this should be "prime factors". You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

Sources

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $6$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $6$