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 $$m.$$MathematicsRS AgarwalStandard X

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