Subjective Type

Show that the square of any positive integer cannot be of the form $$ (6m + 2) $$ or $$ (6m + 5) $$ for any integer $$ m $$ .

Solution

By Euclid's division algorithm , we have
$$ a = 6q + r $$ , where , $$ 0 \leq r < 6 $$
or $$ r = 0 , 1 , 2 , 3 , 4 , 5 $$
Now squaring both sides of $$ a = 6q + r $$
$$ \Rightarrow \, a^3 = (6q)^2 + (r(^2 + 2(6q)(r) $$
$$ \Rightarrow \, a^2 = 6[6q^2 + 2qr] + r^2 $$ .....(i)

At $$ r = 0 $$
$$ a^2 = 6[6q^2 + 2q \times 0] + 0^2 $$ [From (i)]
$$ \Rightarrow \, a^2 = 36q^2 $$
$$ \Rightarrow \,a^2 = 6m $$ , where $$ m = 6q^2 $$

At $$ r = 1 $$
$$a^2 = 6[6q^2 + 2q \times 1] + 1^2 $$ [From (i)]
$$ \Rightarrow \, a^2 = 6[6q^2 + 2q] + 1 $$
$$ \Rightarrow \, a^2 = 6m + 1 $$ , where $$ m = 6q^2 + 2q $$

At $$ r = 2 $$
$$a^2 = 6[6q^2 + 2q \cdot 2] + 2^2 $$ [From(i)]
$$ \Rightarrow \, a^2 = 6m + 4 $$, where $$ m = (6q^2 + 4q) $$

At $$ r = 3 $$
$$ a^2 = 6[6q^2 + 2q \cdot 3] + 3^2 $$ [From (i)]
$$ \Rightarrow \, a^2 = 6[6q^2 + 6q] + 6 + 3 $$
$$ \Rightarrow \, a^2 = 6[6q^2 + 6q + 1] + 3 $$
$$ \Rightarrow \,a^2 = 6m + 3 $$ , where $$ m = 6q^2 + 6q + 1 $$

At $$ r = 4 $$
$$a^2 = 6[6q^2 + 2q \cdot 4] + 4^2 $$
$$ \Rightarrow \, a^2 = 6[6q^2 + 8q] + 12 + 4 $$
$$ \Rightarrow \, a^2 = 6[6q^2 + 8q + 2] + 4 $$
$$ \Rightarrow \, a^2 = 6m + 4 $$ , where $$ m = 6q^4 + 8q + 2 $$

At $$ r = 5 , a^2 = 6[6q^2 + 2q \cdot 5 ] + 5^2 $$ [From(i)]
$$ \Rightarrow \,a^2 = 6[6q^2 + 10q] + 24 + 1 $$
$$ = 6[6q^2 + 10q + 4] + 1 $$
$$ \Rightarrow \,a^2 = 6m + 1 $$, where $$ m = 6q^2 + 10q + 4 $$

Hence , the number of the form $$ 6m , (6m + 1) , (6m + 3) $$ and $$ (6m + 4) $$ are perfect squares and $$ (6m + 2) $$ and , $$ (6m + 5) $$ are not perfect square for some value of $$m$$.

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