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Kaya Renger Coordinator

05/11/2017 at 10:53
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Prove that with n is natural number that :

a) a2 - a \(⋮\) 2    ;    a3 - a \(⋮\) 3     ;     a4 - a \(⋮\) 4

b)  a3 - a \(⋮\) 6    ;    a3 - 7a \(⋮\)  6    ;    a3 + 11a \(⋮\)  6

 




    List of answers
  • ...
    Dao Trong Luan Coordinator 05/11/2017 at 12:01

    a,

    \(a^2-a=a\left(a-1\right)\)

    \(\left[{}\begin{matrix}a⋮2\Rightarrow a\left(a-1\right)⋮2\Rightarrow a^2-a⋮2\\a⋮̸2\Rightarrow a-1⋮2\Rightarrow a\left(a-1\right)⋮2\Rightarrow a^2-a⋮2\end{matrix}\right.\)

    So a2 - a⋮2

    \(a^3-a=a\left(a^2-1\right)\)

    \(\left[{}\begin{matrix}a⋮3\Rightarrow a\left(a^2-1\right)⋮3\Rightarrow a^3-a⋮3\\a⋮3\Rightarrow a^2\equiv1\left(mod3\right)\Rightarrow\left(a^2-1\right)⋮3\Rightarrow a\left(a^2-1\right)⋮3\Rightarrow a^3-a⋮3\end{matrix}\right.\)So a3 - a⋮3

    \(a^4-a=a\left(a^3-1\right)=a\left(a-1\right)\left(a^2+a+1\right)=2k\cdot\left[a\left(a+1\right)+1\right]=2k\left(2h+1\right)=4kh+2k\)So \(a^4-a⋮4\) when a = 4k+1 means is don't always

    b, 

    \(a^3-a⋮3\left(proof-on\right)\)

    \(a^3-a=a\left(a^2-1\right)\)

    \(\left[{}\begin{matrix}a⋮2\Rightarrow a\left(a^2-1\right)⋮2\Leftrightarrow a^3-a⋮2\\a⋮̸2\Rightarrow a^2⋮̸2\Rightarrow\left(a^2-1\right)⋮2\Rightarrow a\left(a^2-1\right)⋮2\Leftrightarrow a^3-a⋮2\end{matrix}\right.\)

    So a3 -a⋮2

    But \(\left(2,3\right)=1\Rightarrow a^3-a⋮6\)

    a3 - 7a = a3 - a - 6a = \(\left(a^3-a\right)-6a=6k-6a=6\left(k-a\right)⋮6\) - Because a3-a⋮6

    So a3-7a ⋮6

    \(a^3+11a=\left(a^3-a\right)+12a=6k+12a=6\left(k+2a\right)⋮6\)

    So a3+11a ⋮ 6

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  • ...
    FA KAKALOTS 09/02/2018 at 22:17

    a,

    a2−a=a(a−1)

    ⎡⎢⎣a⋮2⇒a(a−1)⋮2⇒a2−a⋮2a⋮̸2⇒a−1⋮2⇒a(a−1)⋮2⇒a2−a⋮2

    So a2 - a⋮2

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮3⇒a(a2−1)⋮3⇒a3−a⋮3a⋮3⇒a2≡1(mod3)⇒(a2−1)⋮3⇒a(a2−1)⋮3⇒a3−a⋮3

    So a3 - a⋮3

    a4−a=a(a3−1)=a(a−1)(a2+a+1)=2k⋅[a(a+1)+1]=2k(2h+1)=4kh+2k

    So a4−a⋮4

     when a = 4k+1 means is don't always

    b, 

    a3−a⋮3(proof−on)

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮2⇒a(a2−1)⋮2⇔a3−a⋮2a⋮̸2⇒a2⋮̸2⇒(a2−1)⋮2⇒a(a2−1)⋮2⇔a3−a⋮2

    So a3 -a⋮2

    But (2,3)=1⇒a3−a⋮6

    a3 - 7a = a3 - a - 6a = (a3−a)−6a=6k−6a=6(k−a)⋮6

     - Because a3-a⋮6

    So a3-7a ⋮6

    a3+11a=(a3−a)+12a=6k+12a=6(k+2a)⋮6

    So a3+11a ⋮ 6

    a,

    a2−a=a(a−1)

    ⎡⎢⎣a⋮2⇒a(a−1)⋮2⇒a2−a⋮2a⋮̸2⇒a−1⋮2⇒a(a−1)⋮2⇒a2−a⋮2

    So a2 - a⋮2

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮3⇒a(a2−1)⋮3⇒a3−a⋮3a⋮3⇒a2≡1(mod3)⇒(a2−1)⋮3⇒a(a2−1)⋮3⇒a3−a⋮3

    So a3 - a⋮3

    a4−a=a(a3−1)=a(a−1)(a2+a+1)=2k⋅[a(a+1)+1]=2k(2h+1)=4kh+2k

    So a4−a⋮4

     when a = 4k+1 means is don't always

    b, 

    a3−a⋮3(proof−on)

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮2⇒a(a2−1)⋮2⇔a3−a⋮2a⋮̸2⇒a2⋮̸2⇒(a2−1)⋮2⇒a(a2−1)⋮2⇔a3−a⋮2

    So a3 -a⋮2

    But (2,3)=1⇒a3−a⋮6

    a3 - 7a = a3 - a - 6a = (a3−a)−6a=6k−6a=6(k−a)⋮6

     - Because a3-a⋮6

    So a3-7a ⋮6

    a3+11a=(a3−a)+12a=6k+12a=6(k+2a)⋮6

    So a3+11a ⋮ 6

    a,

    a2−a=a(a−1)

    ⎡⎢⎣a⋮2⇒a(a−1)⋮2⇒a2−a⋮2a⋮̸2⇒a−1⋮2⇒a(a−1)⋮2⇒a2−a⋮2

    So a2 - a⋮2

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮3⇒a(a2−1)⋮3⇒a3−a⋮3a⋮3⇒a2≡1(mod3)⇒(a2−1)⋮3⇒a(a2−1)⋮3⇒a3−a⋮3

    So a3 - a⋮3

    a4−a=a(a3−1)=a(a−1)(a2+a+1)=2k⋅[a(a+1)+1]=2k(2h+1)=4kh+2k

    So a4−a⋮4

     when a = 4k+1 means is don't always

    b, 

    a3−a⋮3(proof−on)

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮2⇒a(a2−1)⋮2⇔a3−a⋮2a⋮̸2⇒a2⋮̸2⇒(a2−1)⋮2⇒a(a2−1)⋮2⇔a3−a⋮2

    So a3 -a⋮2

    But (2,3)=1⇒a3−a⋮6

    a3 - 7a = a3 - a - 6a = (a3−a)−6a=6k−6a=6(k−a)⋮6

     - Because a3-a⋮6

    So a3-7a ⋮6

    a3+11a=(a3−a)+12a=6k+12a=6(k+2a)⋮6

    So a3+11a ⋮ 6

    a,

    a2−a=a(a−1)

    ⎡⎢⎣a⋮2⇒a(a−1)⋮2⇒a2−a⋮2a⋮̸2⇒a−1⋮2⇒a(a−1)⋮2⇒a2−a⋮2

    So a2 - a⋮2

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮3⇒a(a2−1)⋮3⇒a3−a⋮3a⋮3⇒a2≡1(mod3)⇒(a2−1)⋮3⇒a(a2−1)⋮3⇒a3−a⋮3

    So a3 - a⋮3

    a4−a=a(a3−1)=a(a−1)(a2+a+1)=2k⋅[a(a+1)+1]=2k(2h+1)=4kh+2k

    So a4−a⋮4

     when a = 4k+1 means is don't always

    b, 

    a3−a⋮3(proof−on)

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮2⇒a(a2−1)⋮2⇔a3−a⋮2a⋮̸2⇒a2⋮̸2⇒(a2−1)⋮2⇒a(a2−1)⋮2⇔a3−a⋮2

    So a3 -a⋮2

    But (2,3)=1⇒a3−a⋮6

    a3 - 7a = a3 - a - 6a = (a3−a)−6a=6k−6a=6(k−a)⋮6

     - Because a3-a⋮6

    So a3-7a ⋮6

    a3+11a=(a3−a)+12a=6k+12a=6(k+2a)⋮6

    So a3+11a ⋮ 6

    a,

    a2−a=a(a−1)

    ⎡⎢⎣a⋮2⇒a(a−1)⋮2⇒a2−a⋮2a⋮̸2⇒a−1⋮2⇒a(a−1)⋮2⇒a2−a⋮2

    So a2 - a⋮2

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮3⇒a(a2−1)⋮3⇒a3−a⋮3a⋮3⇒a2≡1(mod3)⇒(a2−1)⋮3⇒a(a2−1)⋮3⇒a3−a⋮3

    So a3 - a⋮3

    a4−a=a(a3−1)=a(a−1)(a2+a+1)=2k⋅[a(a+1)+1]=2k(2h+1)=4kh+2k

    So a4−a⋮4

     when a = 4k+1 means is don't always

    b, 

    a3−a⋮3(proof−on)

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮2⇒a(a2−1)⋮2⇔a3−a⋮2a⋮̸2⇒a2⋮̸2⇒(a2−1)⋮2⇒a(a2−1)⋮2⇔a3−a⋮2

    So a3 -a⋮2

    But (2,3)=1⇒a3−a⋮6

    a3 - 7a = a3 - a - 6a = (a3−a)−6a=6k−6a=6(k−a)⋮6

     - Because a3-a⋮6

    So a3-7a ⋮6

    a3+11a=(a3−a)+12a=6k+12a=6(k+2a)⋮6

    So a3+11a ⋮ 6

    a,

    a2−a=a(a−1)

    ⎡⎢⎣a⋮2⇒a(a−1)⋮2⇒a2−a⋮2a⋮̸2⇒a−1⋮2⇒a(a−1)⋮2⇒a2−a⋮2

    So a2 - a⋮2

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮3⇒a(a2−1)⋮3⇒a3−a⋮3a⋮3⇒a2≡1(mod3)⇒(a2−1)⋮3⇒a(a2−1)⋮3⇒a3−a⋮3

    So a3 - a⋮3

    a4−a=a(a3−1)=a(a−1)(a2+a+1)=2k⋅[a(a+1)+1]=2k(2h+1)=4kh+2k

    So a4−a⋮4

     when a = 4k+1 means is don't always

    b, 

    a3−a⋮3(proof−on)

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮2⇒a(a2−1)⋮2⇔a3−a⋮2a⋮̸2⇒a2⋮̸2⇒(a2−1)⋮2⇒a(a2−1)⋮2⇔a3−a⋮2

    So a3 -a⋮2

    But (2,3)=1⇒a3−a⋮6

    a3 - 7a = a3 - a - 6a = (a3−a)−6a=6k−6a=6(k−a)⋮6

     - Because a3-a⋮6

    So a3-7a ⋮6

    a3+11a=(a3−a)+12a=6k+12a=6(k+2a)⋮6

    So a3+11a ⋮ 6

    a,

    a2−a=a(a−1)

    ⎡⎢⎣a⋮2⇒a(a−1)⋮2⇒a2−a⋮2a⋮̸2⇒a−1⋮2⇒a(a−1)⋮2⇒a2−a⋮2

    So a2 - a⋮2

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮3⇒a(a2−1)⋮3⇒a3−a⋮3a⋮3⇒a2≡1(mod3)⇒(a2−1)⋮3⇒a(a2−1)⋮3⇒a3−a⋮3

    So a3 - a⋮3

    a4−a=a(a3−1)=a(a−1)(a2+a+1)=2k⋅[a(a+1)+1]=2k(2h+1)=4kh+2k

    So a4−a⋮4

     when a = 4k+1 means is don't always

    b, 

    a3−a⋮3(proof−on)

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮2⇒a(a2−1)⋮2⇔a3−a⋮2a⋮̸2⇒a2⋮̸2⇒(a2−1)⋮2⇒a(a2−1)⋮2⇔a3−a⋮2

    So a3 -a⋮2

    But (2,3)=1⇒a3−a⋮6

    a3 - 7a = a3 - a - 6a = (a3−a)−6a=6k−6a=6(k−a)⋮6

     - Because a3-a⋮6

    So a3-7a ⋮6

    a3+11a=(a3−a)+12a=6k+12a=6(k+2a)⋮6

    So a3+11a ⋮ 6

    a,

    a2−a=a(a−1)

    ⎡⎢⎣a⋮2⇒a(a−1)⋮2⇒a2−a⋮2a⋮̸2⇒a−1⋮2⇒a(a−1)⋮2⇒a2−a⋮2

    So a2 - a⋮2

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮3⇒a(a2−1)⋮3⇒a3−a⋮3a⋮3⇒a2≡1(mod3)⇒(a2−1)⋮3⇒a(a2−1)⋮3⇒a3−a⋮3

    So a3 - a⋮3

    a4−a=a(a3−1)=a(a−1)(a2+a+1)=2k⋅[a(a+1)+1]=2k(2h+1)=4kh+2k

    So a4−a⋮4

     when a = 4k+1 means is don't always

    b, 

    a3−a⋮3(proof−on)

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮2⇒a(a2−1)⋮2⇔a3−a⋮2a⋮̸2⇒a2⋮̸2⇒(a2−1)⋮2⇒a(a2−1)⋮2⇔a3−a⋮2

    So a3 -a⋮2

    But (2,3)=1⇒a3−a⋮6

    a3 - 7a = a3 - a - 6a = (a3−a)−6a=6k−6a=6(k−a)⋮6

     - Because a3-a⋮6

    So a3-7a ⋮6

    a3+11a=(a3−a)+12a=6k+12a=6(k+2a)⋮6

    So a3+11a ⋮ 6

    a,

    a2−a=a(a−1)

    ⎡⎢⎣a⋮2⇒a(a−1)⋮2⇒a2−a⋮2a⋮̸2⇒a−1⋮2⇒a(a−1)⋮2⇒a2−a⋮2

    So a2 - a⋮2

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮3⇒a(a2−1)⋮3⇒a3−a⋮3a⋮3⇒a2≡1(mod3)⇒(a2−1)⋮3⇒a(a2−1)⋮3⇒a3−a⋮3

    So a3 - a⋮3

    a4−a=a(a3−1)=a(a−1)(a2+a+1)=2k⋅[a(a+1)+1]=2k(2h+1)=4kh+2k

    So a4−a⋮4

     when a = 4k+1 means is don't always

    b, 

    a3−a⋮3(proof−on)

    a3−a=a(a2−1)

    ⎡⎢⎣a⋮2⇒a(a2−1)⋮2⇔a3−a⋮2a⋮̸2⇒a2⋮̸2⇒(a2−1)⋮2⇒a(a2−1)⋮2⇔a3−a⋮2

    So a3 -a⋮2

    But (2,3)=1⇒a3−a⋮6

    a3 - 7a = a3 - a - 6a = (a3−a)−6a=6k−6a=6(k−a)⋮6

     - Because a3-a⋮6

    So a3-7a ⋮6

    a3+11a=(a3−a)+12a=6k+12a=6(k+2a)⋮6

    So a3+11a ⋮ 6


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