STICHTING
MATHEMATISCH CENTRUM
2e BOERHAAVESTRAAT 49
AMSTERDAM

ZW 1957-002

Voordracht in de serie "Actualiteiten"

E.W. Dijkstra

Zaterdag 26 januari 1957


A method to investigate primality


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1957



Voordracht in de serie "Actualiteiten"
door
E.W. Dijkstra
op
Zaterdag 26 Januari 1957


A method to investigate primality

The method determines the smallest odd prime factor of a number N by testing the remainders left after division by the successive odd numbers 3, 5, ... f

m
-2
, f

m
;
here f

m
is the largest odd number not exceeding N. If none of these remainders vanishes, N is a prime number.

Let f be one of the odd divisors or odd trial divisors. Remainder r

0
and quotient q

0
are defined by the relations

N = r

0
+ f.q

0
,           0 ≤ r

0
< f.

Now q

0
is divided by f + 2, giving

q

0
= r

1
+ (f + 2).q

1
,       0 ≤ r

1
< f + 2 .

Then q

1
is divided by (f + 4) etc. and this process is continued till a quotient (q

n
say) equal to zero is found; r

n
is the last remainder in the sequence unequal to zero. After elimination of the q

i
we get the relations
N = r

0
+ f.r

1
+ f.(f+2)r

2
+ f.(f+2).(f+4)r

3
+ ... + f.(f+2) ... (f+2n-2).r

n
   (1)
and
0 ≤ r

i
< f + 2i
   (2)

Once the sequence r

i
is known for a given value of f, it is easy to compute the corresponding sequence r
*
i
,
defined by the relations (1) and (2) with respect to f
*
 
= f + 2,
as they are expressed in terms of the r

i
by the recurrence relations
b

0
= 0, r
*
i
= r

i
- 2(i+1)r

i+1
- b

i
+ (f
*
 
+2i)b

i+1
          (i = 0, 1, ..., n)
   (3)

The relation corresponding to (1) is satisfied for arbitrary values of the numbers b

i
with i ≥ 1; they are fixed, however, by the relations corresponding to (2)
0 ≤ r
*
i
< f
*
 
+ 2i .
   (2*)

On account of the inequalities (2) and (2*) —and b

0
= 0
— the b

i
satisfy the inequalities
0 ≤ b

i
≤ 2i .
   (4)

We have chosen b

0
= 0.
Then the relations (3) and (2*) with i = 0 determine r
*
0
and b

1
;
once b

1
is known, (3) and (2*) with i = 1 determine r
*
1
and b

2
,
etc. The process is easily programmed. [unreadable] 0, and the inequalities (2*) with i = n are always satisfied with b

n+1
= 0,
the process terminates with
r
*
n
= r

n
- b

n

As soon as r
*
n
= 0
is found —in that case it can be proved, that r
*
n-1
≠ 0—
the index n, marking the last r

i
≠ 0
in the sequence, is lowered by 1.

In order to find the smallest odd prime factor of N, the r

i
defined by (2) and (3) and f = 3 are computed. Here the only divisions in the process are carried out. At the same time the initial value of n is found. If N is large, this value may be considerable: for instance n = 11 is found for N ≈ 1013. The amount of work involved in each step is roughly proportional to n2. Fortunately large initial values of n decrease very rapidly. As soon as f.(f + 2).(f + 4) > N, n takes the value 2. This is its minimum value: when r
*
n
= 0
with n = 2 is found (f
*
 
+ 2)2 > N
holds and N is a prime number. (If not, we should have found a r

0
= 0
earlier and should have stopped there.)

The process still may be speeded up. Let b
'
n
be the minimum of b

n
for fixed n up till a certain moment: then it can be shown that the next b

n
satisfies
b

n
b
'
n
+ 1 .
   (5)

Let us apply this to the last stage n = 2. According to (4) b

2
satisfies 0 ≤ b

2
≤ 4.
According to (5), however, the only possible values for b

2
are 0 and 1 as soon as a value b

2
= 0
once has been found. This is bound to happen for f ranging (roughly) from (4N)
1/3
 
till (8N)
1/3
 
.
In the case b

2
= 0
it is apparently unnecessary to test whether r

2
= 0
is reached. (If N ≥ 144, the case b

n
= 0
with n = 2 occurs, before r
*
n
= 0
with n = 2 is found: prime numbers are then always detected in this last stage.)

The less efficient steps of the process for large n (i.e. small f) could be avoided by carrying out divisions for small values of f. (Cf. the method suggested by G.G. Alway, MTAC v. 6, p. 59-60, where one seems to be obliged to do this for f up to (8N)
1/3
 
.)
However we strongly advise not to do this.

If the process described above is started at f = 3, the whole computation can be checked at the end by inserting the final values of f and r

i
into (1). As all the intermediate results are used in the computation, this check seems satisfactory.

If a double length number N is to be investigated, another argument can be added: division of N by small f may give a double length quotient, i.e. two divisions (and two multiplications to check) are needed for each f. In our case even only part of the initial n divisions are double length divisions.

If some consecutive odd numbers are to be investigated, the initial divisions are only necessary for the first number N; it is easy to compute the sequence of initial remainders r

i
"representing N + d at f = 3" from the corresponding sequence for N at f = 3, if d is small.

The process described above has been programmed for the ARMAC (Automatische Rekenmachine van het MAthematisch Centrum). The speed of this machine is about 2400 operations per second. A twelve decimal number was identified as the square of a prime in less than 23 minutes.