Question

show that the square of an odd positive integer is of the form 8m + 1 for some whole number m

Answer 1

hey...here is the answer
Let. .....x=8x+1
squaring both the side's
${(x)}^{2} = {(8x} + {1)}^{2}$
${x}^{2} = {(8x)}^{2} + {(1)}^{2} + 2 \times 8x \times 1$
${x}^{2} = 64 {x}^{2} + 1 + 16x$
${x}^{2} = 8(8 {x}^{2} + {2x}^{2} ) + 1$
Substitute
${(8x}^{2} + 2x)to \: m$
therefore,
${x}^{2} = 8m + 1$
Hence ....we got the answer