Question

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

Answer 1

x^2 + y^2

( x+y)^2 - 2xy

As x and y are both odd integers

so x+y would be even

so let x+y = 2k

(2k)^2 - 2xy

As x and y are odd so xy wouldn't contain factor 2 so it would be odd

So

4 k^2 - 2xy

2( 2k^2 - xy)

As 2k^2 -xy = odd
because even - odd = odd

so it's 2( 2k^2 -xy) is even as it contains factor 2 but not divisible by 4

Answer 2

let x = 2m+1 & y = 2n+1 , m>=0, n>=0.. m & n are integers

now,
${x}^{2} + {y}^{2} = {(2m + 1)}^{2} + {(2n + 1)}^{2} \\ = (4 {m}^{2} + 4m + 1) + (4 {n + 4n + 1}^{2} ) \\ = 4( {m}^{2} + {n}^{2} + m + n) + 2 \\ = 4k + 2 \\ \\ = > \: remainder \: = 2 \: when \: divided \: by \: 4 \\ => {x}^{2} + {y}^{2}\: is\: not\: divisible\: by \:4. \\ or. \\ \\ {x}^{2} + {y}^{2} = 2(2k + 1) = 2a \: \\ \\ = > it \: is \: an \: even\:no.$