The Julian period, proposed by the celebrated Joseph Scaliger as an universal measure of chronology, is formed by taking the continued product of the three cycles of the sun, of the moon, and of the indiction, and is consequently 28 × 19 × 15 = 7980 years. In the course of this long period no two years can be expressed by the same numbers in all the three cycles. Hence, when the number of any proposed year in each of the cycles is known, its number in the Julian period can be determined by the resolution of a very simple problem of the indeterminate analysis. It is unnecessary, however, in the present case to exhibit the general solution of the problem, because when the number in the period corresponding to any one year in the era has been ascertained, it is easy to establish the correspondence for all other years, without having again recourse to the direct solution of the problem. We shall therefore find the number of the Julian period corresponding to the first of our era.

We have already seen that the year 1 of the era had 10 for its number in the solar cycle, 2 in the lunar cycle, and 4 in the cycle of indiction; the question is therefore to find a number such, that when it is divided by the three numbers 28, 19, and 15 respectively the three remainders shall be 10, 2, and 4.

Let x, y, and z be the three quotients of the divisions; the number sought will then be expressed by 28 x + 10, by 19 y + 2, or by 15 z + 4. Hence the two equations

28 x + 10 = 19 y + 2 = 15 z + 4.

To solve the equations 28 x + 10 = 19 y + 2, or y = x + 9 x + 8
line
19
, let m = 9 x + 8
line
19
, we have then x = 2 m + m - 8
line
9
.
Let m - 8
line
9
= m′; then m = 9 m′ + 8; hence

x = 18 m′ + 16 + m′ = 19 m′ + 16 . . . (1).

Again, since 28 x + 10 = 15 z + 4, we have

15 z = 28 x + 6, or z = 2 x - 2 x - 6
line
15
.
Let 2 x - 6
line
15
= n; then 2 x = 15 n + 6, and x = 7 n + 3 + n
line
2
.
Let n
line
2
= n′; then n = 2 n′; consequently

x = 14 n′ + 3 + n′ = 15 n′ + 3 . . . (2).

Equating the above two values of x, we have

15 n′ + 3 = 19 m′ + 16; whence n′ = m′ + 4 m′ + 13
line
15
.
Let 4 m′ + 13
line
15
= p; we have then
4 m′ = 15 p - 13, and m′ = 4 p - p + 13
line
4
.
Let p + 13
line
4
= p′; then p = 4 p′ - 13;

whence m′ = 16 p′ - 52 - p′ = 15 p′ - 52.

Now in this equation p′ may be any number whatever, provided 15 p′ exceed 52. The smallest value of p′ (which is the one here wanted) is therefore 4; for 15 × 4 = 60. Assuming therefore p′ = 4, we have m′ = 60 - 52 = 8; and consequently, since x = 19 m′ + 16, x = 19 × 8 + 16 = 168. The number required is consequently 28 × 168 + 10 = 4714.

Having found the number 4714 for the first of the era, the correspondence of the years of the era and of the period is as follows: -

Era,

1,

2,

3, ...

x,

Period,

4714,

4715,

4716, ...

4713 + x;

from which it is evident, that if we take P to represent the year of the Julian period, and x the corresponding year of the Christian era, we shall have

P = 4713 + x, and x = P - 4713.

With regard to the numeration of the years previous to the commencement of the era, the practice is not uniform. Chronologists, in general, reckon the year preceding the first of the era -1, the next preceding -2, and so on. In this case

Era,

-1,

-2,

-3, ...

-x,

Period,

4713,

4712,

4711, ...

4714 - x;

whence

P = 4714 - x, and x = 4714 - P.

But astronomers, in order to preserve the uniformity of computation, make the series of years proceed without interruption, and reckon the year preceding the first of the era 0. Thus

Era,

0,

-1,

-2, ...

-x,

Period,

4713,

4712,

4711, ...

4713 - x;

therefore, in this case

P = 4713 - x, and x = 4713 - P.