Regular Numbers and Plimpton 322

There are trigonometric arguments for interpretation of Plimpton 322 and a number-theoretic argument by Neugebauer.

Regular (or $$5$$– smooth or Hamming) numbers are of the form :

$$2^i\cdot3^j\cdot5^k$$,  where $$i,j,k = 0,1,2,…$$

That is, a regular number has only  $$2, 3,$$ or $$5$$ as prime factors.

Note that any number raised to  $$0$$ is equal to $$1$$ and that $$1$$ is the multiplicative identity that leaves multiplication unchanged.

For example, $$12 = 2^2\cdot3^1\cdot5^0$$ and  $$5 = 2^0\cdot3^0\cdot5^1$$

Primitive Pythagorean triples are of the form $$(m^2-n^2, 2mn, m^2+n^2)$$.  We have previously shown that primitive Pythagorean triples range through all the numbers $$m$$ and $$n$$, where $$m$$ and $$n$$ have opposite parity, are coprime and $$m > n$$.  We now look at the triples in the tablet where only regular numbers have been chosen for $$m$$ and $$n$$.  This is the number-theoretic argument of Neugebauer $$(1957)$$.

The Babylonians used base $$60$$ for their number system and the regular numbers are integers divisors of powers of $$60$$ and have a finite representation as fractions.

The order of the rows is determined by the tangents (or secants) listed in the first column of the tablet.   This will not be addressed in this article.

We look at the prime factorisations of the triples in the following table :

$$\begin{array}{r|rrrrrrrrrrr} k&m&n&m^2-n^2&pf&d&2mn&pf&d&m^2+n^2&pf&d\\ \hline 1&12&5&119&7\cdot17&4&120&2^3\cdot3\cdot5&16&169&13^2&3\\ 2&64&27&3367&7\cdot13\cdot37&8&3456&2^7\cdot3^3&32&4825&5^2\cdot193&6\\ 3&75&32&4601&43\cdot107&4&4800&2^6\cdot3\cdot5^2&42&6649&61\cdot109&4\\ 4&125&54&12709&71\cdot179&4&13500&2^2\cdot3^3\cdot5^3&48&18541&18541&2\\ 5&9&4&65&5\cdot13&4&72&2^3\cdot3^2&12&97&97&2\\ 6&20&9&319&11\cdot29&4&360&2^3\cdot3^2\cdot5&24&481&13\cdot37&4\\ 7&54&25&2291&29\cdot79&4&2700&2^2\cdot3^3\cdot5^2&36&3541&3541&2\\ 8&32&15&799&17\cdot47&4&960&2^6\cdot3\cdot5&28&1249&1249&2\\ 9&25&12&481&13\cdot37&4&600&2^3\cdot3\cdot5^2&24&769&769&2\\ 10&81&40&4961&11^2\cdot41&6&6480&2^4\cdot3^4\cdot5&50&8161&8161&2\\ 11&2&1&45&3^2\cdot5&6&60&2^2\cdot3\cdot5&12&75&3\cdot5^2&6\\ 12&48&25&1679&23\cdot73&4&2400&2^5\cdot3\cdot5^2&36&2929&29\cdot101&4\\ 13&15&8&161&7\cdot23&4&240&2^4\cdot3\cdot5&20&289&17^2&3\\ 14&50&27&1771&7\cdot11\cdot23&8&2700&2^2\cdot3^3\cdot5^2&36&3229&3229&2\\ 15&9&5&56&2^3\cdot7&8&90&2\cdot3^2\cdot5&12&106&2\cdot53&4 \end{array}$$

where $$pf$$ is the prime factorisation and $$d$$ is the number of divisors.

Note that row $$11$$ is a scalar multiple of $$(3, 4, 5)$$ and row $$15$$ is a scalar multiple of $$(45, 28, 53)$$ which is $$7$$-smooth as a primitive.  All the rest are primitive.

Factorisation of composite numbers has a long history and many methods.  Trial division by the prime numbers is efficient when one of the factors is relatively small.  Fermat’s factorisation method is efficient when one of the factors is relatively close to the square root of the number being factorised; it relies on the algebraic technique of the difference of squares, that is, $$m^2-n^2 = (m-n)(m+n)$$.  Euler’s factorisation method relies on writing an odd number as a sum of two squares in two different ways; the drawback is whether the primes are of the form $$4k=1$$ or $$4k=3$$.

Smooth numbers are used to find the integer factorisation of large semi-primes in cryptography.

The prime factor algorithm is a fast Fourier transform which uses smooth numbers.

Dixon’s factorisation method makes use of the smooth numbers to find a congruence of squares in the modulo of the integer to be factorised.

For example, using row $$1$$, factorise $$119$$.

The square root of $$119$$ is nearly $$11$$.

We look for square numbers close to the root which are $$5$$– smooth in modulo $$119$$.

We find

$$11^2 \equiv 2\pmod {119} = 2$$

$$12^2\equiv 25\pmod{119} = 5^2$$

$$13^2\equiv 50\pmod{119} = 2\cdot5^2$$

We choose $$2\cdot50 = 2^2\cdot5^2 =(2\cdot5)^2 = 10^2$$ because we need even powers.

Therefore, $$10^2\equiv11^2\cdot13^2\pmod{119}$$.

Now $$11\cdot13 = 143\equiv24\pmod{119}$$ and $$24^2\equiv100\pmod{119}$$.

Therefore, $$10^2\equiv24^2\pmod{119}$$.

gcd $$(24\, – 10, 119) =$$ gcd $$(14, 119) = 7$$.

gcd $$(24 + 10, 119) =$$ gcd $$(34, 119) = 17$$.

Hence, $$119 = 7\cdot17$$.

Dixon’s method has been improved by the quadratic sieve.

The quadratic sieve has been improved by the number field sieve (special or general).

Big-O notation is used for determining how much computer time and memory is needed to run algorithms for data sets of size n.

This the hierarchy :

With thanks to FJ for prompting me to look at the numbers in Plimpton 322.