Show that the square of any positive integers is either of the form 3m or 3m + 1 for some ineteger m.

Question

Show that the square of any positive integers is either of the form 3m or 3m + 1 for some ineteger m.

in progress 0
Samantha 2 weeks 2021-09-04T17:34:13+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-09-04T17:35:19+00:00

    Answer:

    Use Euclid division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m. Let a be any positive integer and b = 3. =) … =) a = 3q or 3q + 1 or 3q + 2 for positive integer q.

    0
    2021-09-04T17:36:03+00:00

    Question :–

    Show that the square of any positive integers is either of the form 3m or 3m + 1 for some ineteger m.

    Solution :–

    Let the required integer be,

     \boxed{ \sf \orange{a = bq + r}}

    ATQ,

     \boxed{ \sf \orange{a = 3q + r}} \:  \boxed{ \sf \orange{r = 012}}

    Squaring both sides,

     \boxed{ \sf  {a}^{2} =  {(3q + r)}^{2}  }

     \implies  {a}^{2}  =  {9q}^{2}  + 6qr +  {r}^{2}

     \rule{200}3

    ⓵ If r = 0

     {a}^{2}  =  {9q}^{2}  + 6q(0) +  {(0)}^{2}

     \implies  {a}^{2}  =  {9q}^{2}  = 3 {(3q}^{2} )

     \implies  {a}^{2}  = 3m

    Where m =  \sf {3q}^{2} is another integers.

     \rule{200}3

    ⓶ If r = 1

     {a}^{2}  =  {9q}^{2}  + 6q(1) +  {(1)}^{2}

     \implies  {9q}^{2}  + 6q + 1

     \implies 3( {3q}^{2}  + 2q) + 1

     \implies 3m + 1

    Where m =  \sf {3q}^{2} + 2q is another integers.

     \rule{200}3

    ⓷ If r = 2

     {a}^{2}  =  {9q}^{2}  + 6q(2) +  {(2)}^{2}

     \implies {9q}^{2}  + 12q + 4

     \implies  {9q}^{2}  + 12q + 3 + 1

     \implies 3( {3q}^{2}  + 4q + 1) + 1

     \implies 3m + 1

    Where m =  \sf {3q}^{2} + 4q + 1 is another integers.

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )