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Taylor - Theoretic Arithmetic.pdf - Platonic Philosophy

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$her I ( e~epo~ &YO) which is demonstrated to take place in these<br />

numbers." And he concludes with informing us, "that he<br />

shall discuss in its proper place what is delivered by the Pythagoreans<br />

relative to this most splendid and elegant theory." Unfortunately,<br />

he has not resumed this subject in the above mentioned<br />

treatise, nor in any work of his that is extant. And the<br />

only writer I am acquainted with, who has written more fully<br />

concerning these numbers, is Ozanam, who in his Mathematical<br />

Recreations p. 15. observes of them as follows: "The two<br />

numbers 220 and 284 are called amicable, because the first 220<br />

is equal to the sum of the aliquot parts of the latter, viz. 1+2<br />

+4+71+ 142=220. And reciprocally the latter 248 is equal<br />

to the sum of the aliquot parts of the former, viz. 1 +2+4+5<br />

+lO+ll+22+44+55+ 110=284.<br />

"To find all the amicable numbers in order, make use of<br />

the number 2, which is of such a quality, that if you take 1<br />

from its triple 6, from its sextuple 12, and from the octodecuple<br />

of its square, viz. from 72, the remainders are the three<br />

prime numbers 5, 11 and 71 ; of which 5 and 11 being multiplied<br />

together, and the product 55 being multiplied by 4, the<br />

double of the number 2, this second product 220 will be the<br />

first of the two numbers we look for. And to find the other<br />

284, we need only to multiply the third prime number 71, by<br />

4, the same double of 2 that we used before.<br />

"To find two other amicable numbers, instead of 2 we make<br />

use of one of its powers that possesses the same quality, such as<br />

its cube 8. For if you subtract an unit from its triple 24, from<br />

its sextuple 48, and from 1152 the octodecuple of its square 64,<br />

the remainders are the three prime numbers viz. 23, 47, 1151;<br />

of which the two first 23, 47 ought to be multiplied together,<br />

and their product 1081 ought to be multiplied by 16 the double<br />

of the cube 8, in order to have 17296 for the first of the<br />

two numbers demanded. And for the other amicable number

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