# Thabit's Rule

## Theorem

Let $n$ be a positive integer such that:

 $\displaystyle a$ $=$ $\displaystyle 3 \times 2^n - 1$ $\displaystyle b$ $=$ $\displaystyle 3 \times 2^{n - 1} - 1$ $\displaystyle c$ $=$ $\displaystyle 9 \times 2^{2 n - 1} - 1$

are all prime.

Then:

$\tuple {2^n a b, 2^n c}$

forms an amicable pair.

## Proof

Let $r = 2^n a b, s = 2^n c$.

Let $\map \sigma k$ denote the sigma function on an integer $k$.

$\map \sigma {2^n} = 2^{n + 1} - 1$
 $\displaystyle \map \sigma a$ $=$ $\displaystyle 3 \times 2^n$ $\displaystyle \map \sigma b$ $=$ $\displaystyle 3 \times 2^{n - 1}$ $\displaystyle \map \sigma c$ $=$ $\displaystyle 9 \times 2^{2 n - 1}$
 $\displaystyle \map \sigma r$ $=$ $\displaystyle \map \sigma {2^n} \map \sigma a \map \sigma b$ $\displaystyle$ $=$ $\displaystyle \paren {2^{n + 1} - 1} \paren {3 \times 2^n} \paren {3 \times 2^{n - 1} }$ $\displaystyle$ $=$ $\displaystyle \paren {2^{n + 1} - 1} \paren {9 \times 2^{2 n - 1} }$

and:

 $\displaystyle \map \sigma s$ $=$ $\displaystyle \map \sigma {2^n} \map \sigma c$ $\displaystyle$ $=$ $\displaystyle \paren {2^{n + 1} - 1} \paren {9 \times 2^{2 n - 1} }$ $\displaystyle$ $=$ $\displaystyle \map \sigma r$

Thus it is seen that:

$\map \sigma r = \map \sigma s$

Now we have:

 $\displaystyle r + s$ $=$ $\displaystyle 2^n a b + 2^n c$ $\displaystyle$ $=$ $\displaystyle 2^n \paren {\paren {3 \times 2^n - 1} \paren {3 \times 2^{n - 1} - 1} + 9 \times 2^{2 n - 1} - 1 }$ $\displaystyle$ $=$ $\displaystyle 2^n \paren {9 \times 2^{2 n - 1} - 3 \times 2^n - 3 \times 2^{n - 1} + 1 + 9 \times 2^{2 n - 1} - 1}$ $\displaystyle$ $=$ $\displaystyle 2^n \paren {2 \times 9 \times 2^{2 n - 1} - 3 \times 2 \times 2^{n - 1} - 3 \times 2^{n - 1} }$ $\displaystyle$ $=$ $\displaystyle 2^{n + 1} \paren {9 \times 2^{2 n - 1} - 9 \times 2^{n - 2} }$ $\displaystyle$ $=$ $\displaystyle 2^{n + 1} \paren {9 \times 2^{2 n - 1} } - 2^{n + 1} \paren {9 \times 2^{n - 2} }$ $\displaystyle$ $=$ $\displaystyle 2^{n + 1} \paren {9 \times 2^{2 n - 1} } - \paren {9 \times 2^{2 n - 1} }$ $\displaystyle$ $=$ $\displaystyle \paren {2^{n + 1} - 1} \paren {9 \times 2^{2 n - 1} }$

and so it is seen that:

$r + s = \map \sigma r = \map \sigma s$

Hence the result, by definition of amicable pair.

$\blacksquare$

## Source of Name

This entry was named for Thabit ibn Qurra.

## Historical Note

Thabit's Rule has been discovered and rediscovered several times during the course of history.

The first person on record to discuss it was Thabit ibn Qurra, in his Book on the Determination of Amicable Numbers.

The Thabit pair $\tuple {17 \,296, 18 \, 416}$ was discovered by Ibn al-Banna' al-Marrakushi, who may well have also rediscovered this rule.

In $1636$ it was rediscovered by Pierre de Fermat, who also rediscovered the Thabit pair $\tuple {17 \,296, 18 \, 416}$.

At around the same time, René Descartes also rediscovered it, and also rediscovered the Thabit pair $\tuple {9 \, 363 \, 584, 9 \, 437 \, 056}$.

These three remain the only known Thabit pairs.