Home page Irreducibles Prime/irreducible associations

The reduction process and its terminology were originally defined in Todd's paper "A problem on arc tangent relations" (American Math. Monthly 56 (1949), pp 517-528). The process is outlined below:

An "irreducible" [I]
is the inverse cotangent of an integer I having the property that
the largest prime factor of (I²+1) is not less than 2I. The
primes associated in this way (actually, in a 1-1 correspondence)
with the irreducibles [1], [2], [4], [5], [6], [9] are respectively
2, 5, 17, 13, 37, 41, and so on. The task of reducing [N] (= arccot
N) to a sum of signed irreducible inverse integer cotangents [I1],
[I2], &c., is equivalent to expressing
(N + ** j**) as a
product of factors of the form (I1 ±

Suppose N = 32085. The initial
residue is (32085 + ** j**).

(1) FIND THE LARGEST PRIME FACTOR
'P' OF THE SQUARED MODULUS (X² + Y²) OF THE RESIDUE
(X ± Y** j**), AND HENCE DETERMINE THE NEXT IRREDUCIBLE
TERM (Note: if X and Y are both odd, a factor 2 is present; otherwise
all of the prime factors are congruent to 1, modulo 4).

Initially the squared modulus of the residue is 32085² + 1 = 1029447226, whose largest prime factor is P=829, which is associated (see link above) with the irreducible term [246] - i.e. I=246 is the least integer for which 829 divides (I² + 1); so a term ±[246] appears in the reduction.

(2) DETERMINE THE SIGN OF THE
IRREDUCIBLE TERM.

Suppose the associated irreducible
term is [I]. The sign of [I] is "-" if (X ± YI),
the real integer obtained by substituting 'I' for *' j'*
in the residue, is divisible by P (or is zero); otherwise it is
"+".

Note: (X ± YI) is the coefficient of

Initially, the residue (X ± Y

(3) CALCULATE THE NEW RESIDUE.

This is obtained by multiplying the current residue (X ±
Y** j**) either by (I +

Initially, the product is (32085 +

The new residue (X ± Y

Repeat steps (1), (2) & (3)
until X or Y becomes zero, or until one of them becomes unity
and the other is irreducible, when the reduction will be complete
- in the latter case that irreducible appears as the final term
in the reduction, its sign being calculated as in (2).

Should X and Y both become unity (with the same, or opposite signs)
a [1] term is present in the reduction - its coefficient has to
be worked out by approximating all the inverse cotangent values
present and ensuring they "balance".

Thus (continuing the above example):

(1) 9521² + 39² = 90650962
= 173.97.73.37.2; P = 173, I = 80.

(2) 9521 + 39.80 = 12641, which is NOT divisible by 173; so next
term is +[80].

(3) (9521 + 39** j**).(80 -

(1) 119² + 1 = 14162 = 97.73.2;
P = 97, I = 22.

(2) 119 - 22 IS divisible by 97 (note "-" because of
sign in residue), so next term is -[22].

(3) (119 - ** j**).(22
+

27 is an irreducible, so [27] appears in the reduction; step (2) determines its sign is "+".

So the complete reduction is [32085] = -[22] + [27] + [80] - [246].

Note 1: The order in which terms
in the reduction are found is here dictated by the sizes of the
associated primes.

Note 2: If the highest prime P (>2) found in any step (1) is
present to a higher power than 1, the corresponding irreducible
term will be multiplied by that power - steps (1) and (2) only
have to be carried out once for that term, but step (3) must be
repeated appropriately.