Question

use Euclid's lemma to show that the square of any positive integer is of the form 3p, 3p+1.

Answer 1

let us take, 'x'= 3q , 3q+1, 3q+2
when, x=3q
x2 = (3q) 2
x2 = 9q2
x2 = 3(3q2)
we see that 3q2= m
so we have done the first equation 3m

when , x=3q+1
x2= (3q+1)2
[since, (a+b)2 = a2+2ab+b2]
x2= 9q+6q+1
x2= 3(3q+2q)+1
in this we see that 3q+2q= m
therefore, this satisfy the equation m+1

Answer 2

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♦ Euclid's Lemma :- " Any two natural number 'a' and 'b' can be bound in a generalized result : [ a = bq + r ] where | a | > | b |, and 0 ≤ r < b 

 ◘ For the given question, we write any number as :
    • a = 3q 
    • a = 3q + 1 
    • a = 3q + 2  

Squaring both sides, we get the three consequences :
  • a² = 9q² = 3( 3q² ) = 3p
  • a² = ( 9q² + 6q + 1 ) = 3( 3q² + 2q ) + 1 = 3p + 1
  • a² = ( 9q² + 12q + 4 ) = 3( 3q² + 4q + 1 ) + 1 = 3p + 1 

Hence, we see, the square of any Z⁺ can take a form ( 3p ) or ( 3p + 1 )

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♦ Peeking at Modular Arithmetic, a method in which only the remainder and the divisor counts, 'a' can be written as { 0 , 1 , 2 }( mod 3 ) 

  • a ≡ { 0 , 1 , 2 }( mod 3 )
=> a² ≡ { 0 , 1 }( mod 3 )
=> Square of any +ve integer is of the form 3p or 3p + 1 ; 

Above is a simple two line proof 

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Hope this helps