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Question

Use Euclids division lemma to show that the square of any positive integer is either of the form $$3m$$ or $$3m + 1$$ for some integer $$m$$.


Solution

Using Euclid division algorithm, we know that $$a = bq + r,$$ $$0 \leq r < b$$ .......(1)

Let $$a$$ be any positive integer, and $$b = 3.$$

After substituting $$b = 3$$ in equation $$(i),$$ $$a = 3q + r$$ where $$0 \leq r < 3$$   $$r = 0, 1, 2/$$

If $$r = 0$$ and $$a = 3q,$$
On squaring we get,
$$a^2 = 3(3q^2)$$
$$a^2 = 3m$$, where $$m=3q^2$$

If $$r = 1$$ and $$a = 3q + 1,$$
On squaring we get, 
$$a^2 = 3(3q^2 + 2q) + 1$$
$$a^2=3m+1$$, where $$m=3q^2+2q$$

If $$r = 2$$ and $$a = 3q + 2,$$
On squaring we get,  
$$a^2 = 3(3q^2 + 4q + 1)+1$$
$$a^2=3m+1$$, where $$m=3q^2+4q+1$$

Hence, proved square of any positive integer is either of the forms $$3m$$ or $$3m + 1$$ for some integer $$m.$$

Mathematics
RS Agarwal
Standard X

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