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 ± j), (I2 ± j), &c. (here 'j' is the imaginary square root of -1). After extracting each factor, the "residue" is a complex number (X ± Yj), with X and Y co-prime, whose squared modulus (X² + Y²) steadily decreases.
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 ± Yj), 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 'j'
in the product (X ± Yj)(I + j).
Initially, the residue (X ± Yj) = (32085
+ j); therefore
(X ± YI) = 32085 + 246 = 32331, which IS divisible by 829,
so the sign of [246] in the reduction is "-".
(3) CALCULATE THE NEW RESIDUE.
This is obtained by multiplying the current residue (X ±
Yj) either by (I + j)
or by (I - j),
depending on whether the sign of the irreducible term is "-"
or "+", respectively, and then removing all common factors
(which are certain to include P) from the real and imaginary coefficients
in the product.
Initially, the product is (32085 + j).(246 + j)
= {(32085.246 - 1) + (32085 + 246)j} = {7892909
+ 32331j} = {829.9521 + 829.39j}.
The new residue (X ± Yj) is therefore (9521
+ 39j).
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 + 39j).(80 - j) = (173.37.119 - 173.37j)
so new residue is (119 - j).
(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
+ j) = (97.27
+ 97j) so new residue is (27 + j).
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.