 # Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. Harshit Singh
2 years ago
Dear Student

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,

x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2

Now as per the question given, by squaring both the sides, we get, x2=(3q)^2=9q^2=3×3q^2
Let 3q^2= m
Therefore, x^2= 3m ..........................(1)

x^2= (3q + 1)2= (3q)^2+12+2×3q×1 = 9q^2+ 1 +6q = 3(3q^2+2q) +1 Substitute, 3q^2+2q = m, to get,
x^2= 3m + 1 .................................. (2)
x^2=(3q+2)^2=(3q)^2+22+2×3q×2=9q^2+4+12q=3(3q^2+4q+1)+1
Again, substitute, 3q^2+4q+1 = m, to get,
x^2= 3m + 1................................. (3)
Hence, from equation 1, 2 and 3, we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Thanks
2 years ago

Let a=3b+r ( a, b are positive integers)

The remainder r can be 0,1 or 2

So r = 0,1,2,

Case 1:If r = 0

Then a = 3b

On  squaring we get

a2 = (3b)2

a2 = 9b2

a2 = 3 × 3b2

a2 = 3m ( m=b² is a positive intger)

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Case 2: r = 1

a = 3b + 1

On Squaring we get

a2 = (3b + 1)2

a2 = (3b)2  + 2 × (3b) × 1 + 1

a2 = 9b2 + 6b + 1

a2 = 3(3b2 + 2b) + 1

a2 = 3m + 1

(Here m = 3b2 + 2b is an integer)

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Case 3: r=2

Then a = 3b + 2

On quaring  we get

a2 = (3b + 2)2

a2 = 9b2 + (2 × 3b × 2) +4

a2 = 9b2 + 12b + 3 + 1

a2 = 3(3b2 + 4b + 1) + 1

a2 = 3m + 1

Here m = 3b2 + 4b + 1 is a positive integer

Hence we can conclude that square of any positive integer is either in the

form of 3m or 3m+1( m being a positive integer)