Show that the square of any positive integer cannot be of form 5q+2 or

1.	Show that the square of any positive integer cannot be of form 5q+2 or 5q +3 for any integer q
Answer» Let n be any positive integer. Applying Euclids division lemma with divisor = 5, we get{tex}\\style{font-family:Arial}{\\begin{array}{l}n=5q+1,5q+2,5q+3\\;and\\;5q+4\\;\\\\\\end{array}}{/tex}Now (5q)2 = 25q2 = 5m, where m = 5q2, which is an integer;{tex}\\style{font-family:Arial}{\\begin{array}{l}(5q\\;+\\;1)^{\\;2}\\;=\\;25q^2\\;+\\;10q\\;+\\;1\\;=\\;5(5q^2\\;+\\;2q)\\;+\\;1\\;=\\;5m\\;+\\;1\\\\where\\;m\\;=\\;5q^2\\;+\\;2q,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;2)^2\\;=\\;25q^2\\;+\\;20q\\;+\\;4\\;=\\;5(5q^2\\;+\\;4q)\\;+\\;4\\;=\\;5m\\;+\\;4,\\\\\\;where\\;m\\;=\\;5q^2\\;+\\;4q,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;3)^{\\;2}\\;=\\;25q^2\\;+\\;30q\\;+\\;9\\;=\\;5(5q^2\\;+\\;6q+\\;1)\\;+\\;4\\;=\\;5m\\;+\\;4,\\\\\\;where\\;m\\;=\\;5q^2\\;+\\;6q\\;+\\;1,\\;which\\;is\\;an\\;integer;\\\\\\;(5q\\;+\\;4)^2\\;=\\;25q^2\\;+\\;40q\\;+\\;16\\;=\\;5(5q^2\\;+\\;8q\\;+\\;3)\\;+\\;1\\;=\\;5m\\;+\\;1,\\;\\\\where\\;m\\;=\\;5q^2\\;+\\;8q\\;+\\;3,\\;which\\;is\\;an\\;integer\\\\\\end{array}}{/tex}Thus, the square of any positive integer is of the form 5m, 5m + 1 or 5m + 4 for some integer m.It follows that the square of any positive integer cannot be of the form 5m + 2 or 5m + 3 for some integer m.

Show that the square of any positive integer cannot be of form 5q+2 or 5q +3 for any integer q