Write whether the square of any positive


Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer.


No, by Euclid's lemma, $b=a q+r, 0 \leq r \leq a$ Here, $b$ is any positive integer, $a=3, b=3 q+r$ for $0 \leq r \leq 2$

So, any positive integer is of the form $3 k, 3 k+1$ or $3 k+2$.

$\begin{array}{lll}\text { Now, } & (3 k)^{2}=9 k^{2}=3 m & \text { [where, } m=3 k^{2} \text { ] }\end{array}$

Now, $(3 k)^{2}=9 k^{2}=3 m$ [where, $m=3 k^{2}$ ]

and $\quad(3 k+1)^{2}=9 k^{2}+6 k+1$

$=3\left(3 k^{2}+2 k\right)+1=3 m+1 \quad\left[\right.$ where $\left.m=3 k^{2}+2 k\right]$

Also $\quad(3 k+2)^{2}=9 k^{2}+12 k+4\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$

$=9 k^{2}+12 k+3+1$

$=3\left(3 k^{2}+4 k+1\right)+1$

$=3 m+1 \quad$ [where, $m=3 k^{2}+4 k+1$ ]

which is in the form of $3 m$ and $3 m+1$. Hence, square of any positive number cannot be of the form $3 m+2$.


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