Question

prove that if x and y are both odd positive integers then x2and y2 is even but not divisble by 4

Answer 1

Let the two odd positive numbers be x = 2k + 1 a nd y = 2p + 1
Hence x2 + y2 = (2k + 1)2 + (2p + 1)2
                     = 4k2 + 4k + 1 + 4p2 + 4p + 1
                     = 4k2 + 4p2 + 4k + 4p + 2
                     = 4(k2 + p2 + k + p) + 2
Clearly notice that the sum of square is even the number is not divisible by 4
Hence if x and y are odd positive integers, then x2 + y2 is even but not divisible by 4

Answer 2

___________________________________________________________

♦ ODD NUMBERS CAN BE WRITTEN AS :
         → [ 2k + 1 ] → where 'k' is any WHOLE NUMBER

◙ Let "X" = [ 2a + 1 ]
   Let "Y" = [ 2b + 1 ]   

=> X² = [ 4a² + 4a + 1 ]
     Y² = [ 4b² + 4b + 1 ]

• Adding up the two :
  → [ X² + Y² ] = [ 4( a² + b² ) + 4( a + b ) + 2 ] 
                       = 2[ 2( a² + b² ) + 2( a + b ) + 1 ]
_____________________________________________________________

=> [ X² + Y² ] ≡ 2M => [ X² + Y² ] is even 
_____________________________________________________________

• Further, 
    [ X² + Y² ] = [ 4( a² + b² ) + 4( a + b ) + 2 ] 
                    = 4[ a² + b² + a + b ] + 2

→ Comparing this to Euclid's Lemma, one can easily determine that the remainder when [ X² + Y² ] is divided by '4' ≠ 0, and hence, it is not divisible by '4'
_____________________________________________________________

→ Using Modular Arithmetic would be even great :
  → [ X , Y ] ≡ 1 ( mod 2 ) 
=>  [ X² + Y² ] ≡ ( 1 + 1 ) ≡ 0 ( mod 2 )
 
• Hence, EVEN 
______________________________________________________
 
  → Odd squares ≡ 1 ( mod 4 )
      => [ X² + Y² ] ≡ 2 ( mod 4 )

• Hence, not divisible by 4
_______________________________________________________

♦ HOPE THIS HELPS ^_^