第9卷命题 36 · 完全数定理

For let as many numbers as we please, A, B, C, D, beginning from an unit be set out in double proportion, until the sum of all becomes prime, let E be equal to the sum, and let E by multiplying D make FG; I say that FG is perfect. For, however many A, B, C, D are in multitude, let so many E, HK, L, M be taken in double proportion beginning from E; therefore, ex aequali, as A is to D, so is E to M. [VII. 14] Therefore the product of E, D is equal to the product of A, M. [VII. 19] And the product of E, D is FG; therefore the product of A, M is also FG. Therefore A by multiplying M has made FG; therefore M measures FG according to the units in A. And A is a dyad; therefore FG is double of M. But M, L, HK, E are continuously double of each other; therefore E, HK, L, M, FG are continuously proportional in double proportion. Now let there be subtracted from the second HK and the last FG the numbers HN, FO, each equal to the first E; therefore, as the excess of the second is to the first, so is the excess of the last to all those before it.

设从单位开始成双倍比例的数A、B、C、D，其和E为素数，E乘以D得FG。取与A、B、C、D个数相同的数E、HK、L、M，也成双倍比例。由等比性质，A比D等于E比M，故E乘D等于A乘M，即FG等于A乘M，因此M量尽FG，且A是2，故FG是M的两倍。

[IX. 35] Therefore, as NK is to E, so is OG to M, L, KH, E. And NK is equal to E; therefore OG is also equal to M, L, HK, E. But FO is also equal to E, and E is equal to A, B, C, D and the unit. Therefore the whole FG is equal to E, HK, L, M and A, B, C, D and the unit; and it is measured by them. I say also that FG will not be measured by any other number except A, B, C, D, E, HK, L, M and the unit. For, if possible, let some number P measure FG, and let P not be the same with any of the numbers A, B, C, D, E, HK, L, M. And, as many times as P measures FG, so many units let there be in Q; therefore Q by multiplying P has made FG. But, further, E has also by multiplying D made FG; therefore, as E is to Q, so is P to D.

从HK和FG中分别减去等于E的HN和FO，则第二个的剩余NK比E等于最后一个的剩余OG比它前面的所有数M、L、KH、E。因NK等于E，故OG等于M、L、HK、E之和。又FO等于E，E等于单位加A、B、C、D，所以FG等于E、HK、L、M、A、B、C、D及单位之和，且被它们量尽。

[VII. 19] And, since A, B, C, D are continuously proportional beginning from an unit, therefore D will not be measured by any other number except A, B, C. [IX. 13] And, by hypothesis, P is not the same with any of the numbers A, B, C; therefore P will not measure D. But, as P is to D, so is E to Q; therefore neither does E measure Q. [VII. Def. 20] And E is prime; and any prime number is prime to any number which it does not measure. [VII. 29] Therefore E, Q are prime to one another. But primes are also least, [VII. 21] and the least numbers measure those which have the same ratio the same number of times, the antecedent the antecedent and the consequent the consequent; [VII. 20] and, as E is to Q, so is P to D; therefore E measures P the same number of times that Q measures D. But D is not measured by any other number except A, B, C; therefore Q is the same with one of the numbers A, B, C. Let it be the same with B.

假设另有数P量尽FG，且P不同于A、B、C、D、E、HK、L、M。设Q次量尽，则Q乘P等于FG，而E乘D也等于FG，故E比Q等于P比D。因A、B、C、D从单位起成比例，D只能被A、B、C量尽，而P不是它们之一，故P不量尽D，从而E不量尽Q。

And, however many B, C, D are in multitude, let so many E, HK, L be taken beginning from E. Now E, HK, L are in the same ratio with B, C, D; therefore, ex aequali, as B is to D, so is E to L. [VII. 14] Therefore the product of B, L is equal to the product of D, E. [VII. 19] But the product of D, E is equal to the product of Q, P; therefore the product of Q, P is also equal to the product of B, L. Therefore, as Q is to B, so is L to P. [VII. 19] And Q is the same with B; therefore L is also the same with P; which is impossible, for by hypothesis P is not the same with any of the numbers set out. Therefore no number will measure FG except A, B, C, D, E, HK, L, M and the unit.

E是素数，与Q互素，故E与Q是最小的。由比例性质，E量尽P的次数等于Q量尽D的次数。但D只能被A、B、C量尽，故Q等于其中之一，设为B。取与B、C、D个数相同的E、HK、L，则B比D等于E比L，故B乘L等于D乘E，也等于Q乘P。因Q等于B，故L等于P，矛盾。因此只有所列数及单位能量尽FG。