Question

Prove that the square of any integer is of the form 5m, 5m+1, 5m+4

Answer 1

Sol :
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 iseither of the form 5n or 5n+1 for integer n.

Answer 2

Hey! ! !

Answer :-

☆ Let x be any positive integer


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


If x = 5q, x2 = (5q)2 = 25q2 = 5(5q2) = 5n (where n = 5q2 )
If x = 5q+1, x2 = (5q+1)2 = 25q2+10q+1 = 5(5q2+2q)+1 = 5n+1 (where n = 5q2+2q )
If x = 5q+4, x2 = (5q+4)2 = 25q2+40q+16 = 5(5q2 + 8q + 3)+ 1 = 5n+1 (where n = 5q2+8q+3 )


∴in each of three cases x2 is either of the form 5q or 5q+1 or 5q+4 and for integer q

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