60 . AN ESSAY ON THE ROOTS OF INTEGERS, 



(17). Now let the highest approximate 6th Root of A be a, so that 

 a 6 Z, and (a -f l) 6 > A. Then is a a Digit. If not a = or > p. 



First. Let a ~ p, then a 6 — p 6 and contains 7 figures by Lem. 2, 

 and yet A only contains 3 figures, which is absurd. A fortiori a cannot 

 be > p. 



(18). Let A— a 6 = R, then R Z_ 6 a 5 -f 15 a 4 -f- 20 a s -f 15 a 2 -f- 6 a 

 -f 1. If not R = or > 6 a 5 -f- 15 a 4 -f 20 a 3 -f 15 a 2 -f 6 a -f 1. 



First, let it be equal. Then A — a 6 — 6 a s -f 15 « 4 -f 20 a s -J- 15 a s 

 -f 6 a -f 1, and A = a 6 -f- 6 a 5 -f 15 a 4 + 20 a 3 -f 15 a 2 + 6 a -|- 1 = 

 (a -f l) 6 . And yet by supposition A Z. ( a+ \f — which is absurd. 

 Second, a fortiori it cannot be greater. 



(19.) Since A— a 6 = R and A = « 6 -f- R, so A p 6 -f B = (a 6 -f R) p 9 



■f B = a 6 p 6 -f R p 6 -f B = a <p [ 6 -f (R p 6 -f B). Now since 



\ a p | = a p so by Par. 13) a <p will here represent the S and R p 6 -f 

 B the t of that Paragraph and if b be the greatest possible integer such 

 that 6. a p \ 5 b -f 15. "o^Tj 4 & 2 -f 20." a p "| 3 6 3 -f 15. a p J 2 6 4 -J- 

 6. a f J 5 + & 6 that is, 6 a s p s . b -f 15 a 4 p 4 6 2 -f 20 a 3 p 3 Z> 3 -f 15 a 2 p 2 V 

 4- 6 a p b s + b 6 should be less than R p 6 -J- B, then is a p + b the 

 highest approximate 6 Root of a p | 6 -j- R <p 6 -j- B or A p 6 -f B. and b 

 will represent thew of Par. 13). 



(20.) In this case b when found must be a Digit. If not b = or > p. 



First. Let b = p. Then 6 a 5 p s b -f 15 a 4 p 4 Z> 2 -f 20 a 3 p 3 b 3 -f- 

 15 a 2 p 2 6 4 -j- 6 a p b s -j- b 6 = 6 a s <f -f 15 a 4 p° -j- 20 a 3 p 6 + 15 a 2 p 6 -f 

 6a^ + p\ . Now R by Par. 18 z_ 6 a 5 + 15 a 4 -f 20 a 3 -f 15 a 2 -f 



