Question

use euclids division lemma to show that the square of any positive integer is of the form 5n ,5n+1 ,5n+4

Answer 1

Answer:

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Let x be any positive integer Then x = 5q or x = 5q+1 or x = 5q+4  for integer x. If x = 5q, x2 = (5q)2 = 25q2 = 5(5q2) = 5n (where n = 5q2 ) If x = 5q+1, x2 = (5q+1)2 = 25q2+10q+1 = 5(5q2+2q)+1 = 5n+1 (where n = 5q2+2q ) If x = 5q+4, x2 = (5q+4)2 = 25q2+40q+16 = 5(5q2 + 8q + 3)+ 1 = 5n+1 (where n = 5q2+8q+3 ) ∴in each of three cases x2 is either of the form 5q or 5q+1 or 5q+4 and for integer q.

Answer 2

Answer:

let as start taking a where a is an positive integer

we apply division algorithm where a and b=5

since 0 ≤ r <3,

the possible remainders are o,1,2,3,4

it can be written in the form 5q or 5q+1 or 5q+2 or 5q+3 or 5q+4 where q is the quotient

Step-by-step explanation:

(5q)²=25q =5n [since 5 is divisible by 25]

(5q+1)²=25q+1+10q

=5(5q+2q)+1

=5n+1  [5q+2q=n]

(5q+2)²=25q+4+20q

=5[5q+4q]+4

=5n+4

therefore any positive odd integer is of the form 5n,5n+1,5n+4

hence proved