Question

For an integer m, the square of any positive integer cannot be of the form

• A. 6m + 4
• B. 5m + 4
• C. 4m + 1
• D. 6m + 5

Answer 1

The correct option is D 6m + 5

Let a be any positive integer.

By Euclid’s division lemma,

a = bq + r

a = 6q + r (when b = 6)

So, r can be 0, 1, 2, 3, 4, 5.

Case 1:

∴ a = 6q (When r = 0)

$\Rightarrow a^2=36q^2$

$\Rightarrow a^2=6\left(6q^2\right)=6m(\mathrm{Where}m=6q^2)$
Case 2:

$\therefore a=6q+1(\mathrm{When}r=1)$

$\Rightarrow a^2={(6q+1)}^2$

$\Rightarrow a^2=36q^2+12q+1=6(6q^2+2q)+1$

$\Rightarrow a^2=6m+1(\mathrm{Where}m=6q^2+2q)$

Case 3:

∴ a = 6q + 2 (When r = 2)

$\Rightarrow a^2={(6q+2)}^2$

$\Rightarrow a^2=36q^2+24q+4=6(6q^2+4q)+4$

$\Rightarrow a^2=6m+4(\mathrm{Where}m=6q^2+4q)$
Case 4:

∴ a = 6q + 3 (When r = 3)


$\Rightarrow a^2={(6q+3)}^2$

$\Rightarrow a^2=36q^2+36q+9=36q^2+36q+6+3=6(6q^2+6q+1)+3$
$\Rightarrow a^2=6m+4(\mathrm{Where}m=6q^2+8q+2)$

Case 6:

∴ a = 6q + 5 (When r = 5)

$\Rightarrow a^2={(6q+5)}^2$

$\Rightarrow a^2=36q^2+60q+25=36q^2+60q+24+1=6(6q^2+10q+4)+1$

$\Rightarrow a^2=6m+1(\mathrm{Where}m=6q^2+10q+4)$

Similarly, for a = bq + r at b = 5 and r = 2

∴ a = 5q + 2

$\Rightarrow a^2={(5q+2)}^2$

$\Rightarrow a^2=25q^2+20q+4=5(5q^2+4q)+4$

$\Rightarrow a^2=5m+4(\mathrm{Where}m=5q^2+4q)$

and at b = 4 and r = 1

∴ a = 4q + 1

$\Rightarrow a^2={(4q+1)}^2=16q^2+8q+1=4(4q^2+2q)+1$
$\Rightarrow a^2=4m+1(\mathrm{Where}m=4q^2+2q)$

From, these cases, this can be concluded that square of any positive number cannot be of the form 6m + 5.

Hence, the correct answer is option (d).