Let m be a natural number for how many values of m 4m+1 is a perfect s

1.	Let m be a natural number for how many values of m 4m+1 is a perfect square
Answer» Let a be any positive integer.Applying Euclid’s division lemma with divisor = 2, we get{tex}\\begin{array}{l}a=2q+r\\;\\;\\;\\;\\;\\;\\;\\;\\;0\\leq r<2\\\\So\\;r=0,1\\\\\\end{array}{/tex}When r = 0,a = 2qSo a2= (2q)2 = 4q2 = 4m--------(1) ( where\xa0m = q2, which is an integer)When r = 1Then a= 2q+1a2= (2q + 1) 2 = 4q2 + 4q + 1 = 4(q2 + q ) +1 = 4m+1 --------(2) (where m = q2 + q, which is an integer)From (1) and (2) We Can conclude that\xa0The square of any positive integer is of the form 4m or 4m +1 for some integer m.

Discussion