Question

using EDL to show that the square of any positive integer is 5m, 5m+1, 5m+4

Answer 1

$\huge \star {\underline {{SOLUTION }}}$

Let x be any integer

Then x = 5m or x = 5m+1 or x = 5m+4 for integer x.

If x = 5m, x2 = (5m)2 = 25m2 = 5(5m2) = 5n (where n = 5m2 )

If x = 5m+1, x2 = (5m+1)2 = 25m2+10m+1 = 5(5m2+2m)+1 = 5n+1 (where n = 5m2+2m )

If x = 5m+4, x2 = (5m+4)2 = 25m2+40m+16 = 5(5m2+8m+3)+1 = 5n+1 (where n = 5m2+8m+3 )

∴in each of three cases x2 is either of the form 5n or 5n+1 for integer n.

HOPE THIS WILL HELP YOU

Answer 2

Let x be any integer

Then x = 5m or x = 5m+1 or x = 5m+4 for integer x.

If x = 5m, x2 = (5m)2 = 25m2 = 5(5m2) = 5n (where n = 5m2 )

If x = 5m+1, x2 = (5m+1)2 = 25m2+10m+1 = 5(5m2+2m)+1 = 5n+1 (where n = 5m2+2m )

If x = 5m+4, x2 = (5m+4)2 = 25m2+40m+16 = 5(5m2+8m+3)+1 = 5n+1 (where n = 5m2+8m+3 )

∴in each of three cases x2 is either of the form 5n or 5n+1 for integer n.