Question

Show that the square of any positive integer cannot be of the form 3m+2, where m is a natural number

Answer 1

Justification:Let a be any positive integer. Then by Euclid’s division lemma, we havea = bq + r, where 0 ≤ r < bFor b = 3, we havea = 3q + r, where 0 ≤ r < 3 ...(i)So, The numbers are of the form 3q, 3q + 1 and 3q + 2.So, (3q)2 = 9q2 = 3(3q2)= 3m, where m is a integer.(3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1= 3m + 1,where m is a integer.(3q + 2)2 = 9q2 + 12q + 4,which cannot be expressed in the form 3m + 2.Therefore, Square of any positive integer cannot be expressed in the form 3m + 2.

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Answer 2

$\huge \fbox \red{Solution:}$

$\rightarrow \rm \blue{Let \: the \: positive \: integer \: be \: ‘a’}$

$\fbox \orange{According to Euclid's division lemma,}$

$a = bm + r$$\rightarrow \rm \blue{According \: to \: the question, we \: take \: } \\ \rm\blue{ b = 3}$

a = 3m + r

So, r = 0, 1, 2.

When r = 0, a = 3m.

When r = 1, a = 3m + 1.

When r = 2, a = 3m + 2.

Now,

When a = 3m

[tex] {a}^{2}

= {3m}^{2}

= 9m2

a

2

= 3(3m2

) = 3q, where q = 3m2[/tex]