Question

Prove that if x and y are both odd positive integers, then x 2 + y 2 is even but not divisible by 4.

Answer 1

Answer:

let the two Odd number be(2a+1)&(2b+1) because if we add 1to any even number it will be Odd

Answer 2

Appropriate Question:

$\sf{Prove \: that \: if \: x \: and \: y \: are \: both \: odd \: positive \: integers, \: then \: {x}^{2} + {y}^{2} \: is \: even \: but \: not \: divisible \: by \: 4.}$

Solution:

We know that any odd positive integer is of form 2q + 1 for some integer q.

Consider x = 2m + 1 and y = 2n + 1 for some integer m and n.

$\sf{ \therefore \: \: \: \: \: \: \: {x}^{2} + {y}^{2} = (2m + 1 )^{2} + {(2n + 1)}^{2} } \\ : \implies \sf{ {x}^{2} + {y }^{2} = {4m}^{2} + 1 + 4m + {4n}^{2} + 1 + 4n } \\ : \implies \sf{ {x}^{2} + {y}^{2} = {4m}^{2} + {4n}^{2} + 4m + 4n + 2} \\ : \implies \sf{ {x}^{2} + {y}^{2} = 4 \{( {m}^{2} + {n}^{2} ) + (m + n) \} + 2 } \\ : \implies \sf{ {x}^{2} + {y}^{2} = 4q + 2 } \\ \sf{where \: q = ( {m}^{2} + {n}^{2} ) + (m + n)}$

$: \implies \sf{ {x}^{2} + {y}^{2} \: is \: even \: and \: leaves \: remainder \: 2 \: when \: divided \: by \: 4. } \\ \sf{hence \: {x}^{2} + {y}^{2} \: is \: even \: but \: not \: divisible \: by \: 4. }$

Hence Proved.