Kaya Renger Coordinator
05/11/2017 at 10:53-
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
Selected by MathYouLike -
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