Equating the above two values of x, we have
| 15 n′ + 3 = 19 m′ + 16; whence n′ = m′ + | 4 m′ + 13 15 | . |
| Let | 4 m′ + 13 15 | = p; we have then |
| 4 m′ = 15 p - 13, and m′ = 4 p - | p + 13 4 | . |
| Let | p + 13 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