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Home » 12. Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.
12. Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.
Class 10 | Maths | RD Sharma | Real Numbers
12. Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Solution:

Let the positive integer = a

According to Euclid’s division algorithm,

a = 6q + r, where 0 ≤ r < 6 

    \[{{a}^{2}}={{\left( 6q+r \right)}^{2}}=36{{q}^{2}}+{{r}^{2}}+12qr\]

,

    \[\because {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]

    \[{{a}^{2}}=6\left( 6q{{r}^{2}}+2qr \right)+{{r}^{2}}\]

…(i), where,0 ≤ r < 6 

When r = 0, substituting r = 0 in Eq.(i), we get 

    \[{{a}^{2}}=6\left( 6{{q}^{2}} \right)=6m\]

, where, m = 6q2 is an integer.

When r = 1, substituting r = 1 in Eq.(i), we get

    \[{{a}^{2}}=\left( 6{{q}^{2}}+2q \right)+1=6m+1\]

, where, m = (6q2 + 2q) is an integer.

When r = 2, substituting r = 2 in Eq(i), we get

    \[{{a}^{2}}=6\left( 6{{q}^{2}}+4q \right)+4=6m+4\]

, where, m =

    \[\left( 6{{q}^{2}}+4q \right)\]

is an integer.
When r = 3, substituting r = 3 in Eq.(i), we get

    \[{{a}^{2}}=6\left( 6{{q}^{2}}+6q \right)+9=6\left( 6{{q}^{2}}+6q \right)+6+3\]

    \[{{a}^{2}}=6\left( 6{{q}^{2}}+6q+1 \right)+3=6m+3\]

, where

    \[m=\left( 6{{q}^{2}}+6q+1 \right)\]

is integer.

When r = 4, substituting r = 4 in Eq.(i) we get 

    \[{{a}^{2}}=6\left( 6{{q}^{2}}+8q \right)+16=6\left( 6{{q}^{2}}+8q \right)+12+4\]

    \[{{a}^{2}}=6\left( 6{{q}^{2}}+8q+2 \right)+4=6m+4\]

, where, m

    \[m=\left( 6{{q}^{2}}+6q+1 \right)\]

is an integer.

When r = 5, substituting r = 5 in Eq.(i), we get

    \[{{a}^{2}}=6\left( 6{{q}^{2}}+10q \right)+25=6\left( 6{{q}^{2}}+10q \right)+24+1\]

    \[{{a}^{2}}=6\left( 6{{q}^{2}}+10q+4 \right)+1=6m+1\]

, where, m

    \[m=\left( 6{{q}^{2}}+10q+4 \right)\]

is an integer.

Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m. 

Hence Proved.

i

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