show that the square of any positive integer is the form 4m or 4m+1 for some integer m.
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let a= any positive integer
b= 4
r= 0,1,2,3
According to Euclid's division lemmma: a=bq+r
we have,
a= 4q, 4q+1, 4q+2, 4q+3
Case 1: a= 4q
Squaring them we get,
a =(4q)²
a = 16q²
a = 4(4q)
a = 4m
Case 2: a= 4q+1
Squaring them we get,
a =(4q+1)²
a =(4q)² + (1)² + 8q
a =16q² + 8q + 1
a = 4(4q² + 2) + 1
a = 4m+1
Case 3: a= 4q+2
Squaring them we get,
a =(4q+2)²
a =(4q)² + (2)² + 16q
a =16q² + 16q + 4
a = 4(4q² + 4q + 1)
a = 4m
Case 4: a= 4q+3
Squaring them we get,
a =(4q+3)²
a =(4q)² + (3)² + 24q
a =16q² + 24q + 9
a =16q² + 24q + 8 + 1
a = 4(4q² + 8q + 2) + 1
a = 4m+1
therefore, square of any positive integer is in the form 4m or 4m+1
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