Let there be as many numbers as we please, A, B, C, and D, beginning from a unit and in continued proportion, and let A, the number after the unit, be prime.
I say that D, the greatest of them, is not measured by any other number except A, B, or C.
If possible, let it be measured by E, and let E not be the same with any of the numbers A, B, or C.
It is then clear that E is not prime, for if E is prime and measures D, then it also measures A, which is prime, though it is not the same with it, which is impossible. Therefore E is not prime, so it is composite.
But any composite number is measured by some prime number, therefore E is measured by some prime number.
I say next that it is no measured by any other prime except A.
If E is measured by another, and E measures D, then that other measures D, so that it also measures A, which is prime, though it is not the same with it, which is impossible. Therefore [only the prime] A measures E.And, since E measures D, let it measure it according to F.
I say that F is not the same with any of the numbers A, B, or C.
If F is the same with one of the numbers A, B, or C, and measures D according to E, then one of the numbers A, B, or C also measures D according to E. But one of the numbers A, B, or C measures D according to some one of the numbers A, B, or C, therefore E is also the same with one of the numbers A, B or C, which is contrary to the hypothesis.
Therefore F is not the same as any one of the numbers A, B, or C.
Similarly we can prove that F is measured by A, by proving again that F is not prime.
If it is, and measures D, then it also measures A, which is prime, though it is not the same with it, which is impossible. Therefore F is not prime, so it is composite.
But any composite number is measured by some prime number, therefore F is measured by some prime number.
I say next that it is not measured by any other prime except A.
If any other prime number measures F, and F measures D, then that other also measures D, so that it also measures A, which is prime, though it is not the same with it, which is impossible. Therefore [only the prime] A measures F.
And, since E measures D according to F, therefore E multiplied by F makes D.
But, further, A multiplied by C makes D, therefore the product of A and C equals the product of E and F.
Therefore, proportionally A is to E as F is to C.
But A measures E, therefore F also measures C. Let it measure it according to G.
Similarly, then, we can prove that G is not the same with any of the numbers A or B, and that it is measured by A. And, since F measures C according to G, therefore F multiplied by G makes C.
But, further, A multiplied by B makes C, therefore the product of A and B equals the product of F and G. Therefore, proportionally A is to F as G is to B.
But A measures F, therefore G also measures B. Let it measure it according to H.
Similarly then we can prove that H is not the same with A.
And, since G measures B according to H, therefore G multiplied by H makes B. But, further, A multiplied by itself makes B, therefore the product of H and G equals the square on A.
Therefore H is to A as A is to G. But A measures G, therefore H also measures A, which is prime, though it is not the same with it, which is absurd.
Therefore D the greatest is not measured by any other number except A, B, or C.
Therefore, if as many numbers as we please beginning from a unit are in continued proportion, and the number after the unit is prime, then the greatest is not measured by any except those which have a place among the proportional numbers.
Suppose a number e divides a power pk of a prime number p, but e does not equal any lower power of p.
First note that e can’t be prime itself, since then it would divide p (IX.12), which it doesn’t.
Then e is composite. Then some prime number q divides e (VII.31). Then q also divides pk, which it implies q divides p. Therefore, the only prime that can divide e is p.
The rest of the proof is repeated reduction of the power k. Since e is not 1, it is divisible by p. Let g be e/p. Then g divides pk-1, but is not any lower power of p. Then the same argument can be applied. Continue in this manner until some number divides p but is not 1 or p, a contradiction. Thus, the only numbers that can divide a power of a prime are smaller powers of the prime.