Question

1. Prove that difference of squares of two distinct
odd natural numbers is always a multiple of 8.​

Answer 1

$\huge\underline\mathfrak\red{Answer}$

$as \: we \: know \: that \: any \: odd \: no. \: can \\ be \: written \: as \\ \\ 4r + b \: where \: 0 \leqslant b < 4 \\ where \: b \: is \: an \: odd \: no. \: smaller\\ than \: 4 \\ \\ let \: any \: odd \: no. \: x \\ \\ x = 4r + b \\ \\ squaring \\ \\ {x}^{2} = 16 {r}^{2} + {b}^{2} + 8rb......(1) \\ \\ let \: another \: odd \: no. \: y \\ \\ y = 4s+ c \\ \\ squaring \\ \\ {y}^{2} = 16 {s}^{2} + {c}^{2} + 8sc ......(2)\\ \\ where \: s \: \: is \: another \: \: no. \: other \\ than \: r \: and \: c \: i s\: another \: odd \: no. \: \\ equal \: to\: b\\ \\ subtract \: (1) \: from \: (2) \\ \\ {y}^{2} - {x}^{2} = 16 {s}^{2} + {c}^{2} + 8sc - 16 {r}^{2} - {b}^{2} - 8rb \\ \\ {y}^{2} - {x}^{2} = 16( {s}^{2} - {r}^{2} ) + 8(sc - rb) + {c}^{2} - {c}^{2} \\ \\ {y}^{2} - {x}^{2} = 8(2 {s}^{2} - 2 {r}^{2} + sc - rb) \\ \\ let \: m = 2 {s}^{2} - 2 {r}^{2} + sc - rb \\ \\ {y}^{2} - {x}^{2} = 8m$