Question

# Use Euclid's division lemma to show that the square of any positive integer cannot be of the form 5m + 2 or 5m + 3 for some integer m.

Solution

## Let a be any positive integer. By Euclid's division lemma, a = bm + r where b = 5 ⇒ a = 5m + r So, r can be any of 0, 1, 2, 3, 4 ∴ a = 5m + 1 when r = 0 a = 5m + 1 when r = 1 a = 5m + 2 when r = 2 a = 5m + 3 when r = 3 a = 5m + 4 when r = 4 So, 'a' is any positive integer in the form of 5m, 5m + 1, 5m + 2, 5m + 3, 5m + 4 for some integer m. Case I: a = 5m ⇒ a2=(5m)2=25m2⇒ a2=5(5m2) = 5q, where q=5m2 Case II: a = 5m + 1 ⇒ a2=(5m+1)2=25m2+10m+1⇒ a2=5(5m2+2m)+1 = 5q + 1, where q=5m2+2m Case III: a = 5m + 2 ⇒ a2=(5m+1)2=25m2+20m+4=25m2+20m+4=5(5m2+4m)+4 = 5q + 4 where q = 5m2 + 4m Case IV: a = 5m + 3 ⇒a2=(5m+32)=25m2+30m+9=25m2+30m+5+4 = 5 (5m2 + 6m + 1) + 4 = 5q + 4 where q = 5m2 + 6m + 1 Case V: a = 5m + 4 ⇒ a2=(5m+4)2=25m2+40m+16 = 25m2 40m + 15 + 1 = 5(5m2 + 8m + 3) + 1 = 5q + 1 where q = 5m2 + 8m + 3 From all these cases, it is clear that square of any positive integer can not be of the form 5m + 2 or 5m + 3

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