Prove that ## 13 ## divides ## 10^{2p}-10^{p}+1 ##

[Math100]

Homework Statement

For any prime $p>3$, prove that $13$ divides $10^{2p}-10^{p}+1$.

Relevant Equations

None.

Proof:

Let $p>3$ be any prime number.
Then $p>3$ is either of the form $6k+1$ or $6k+5$ for some $k\in\mathbb{Z}$.
Now we consider two cases.
Case #1: Suppose $p=6k+1$ for some $k\in\mathbb{Z}$.
Then $10^{2p}-10^{p}+1=10^{12k+2}-10^{6k+1}+1$.
Observe that
\begin{align*}
&10^{2p}-10^{p}+1\equiv [(-3)^{12k+2}-(-3)^{6k+1}+1]\pmod {13}\\
&\equiv [9((-3)^{3})^{4k}\cdot 3((-3)^{3})^{2k}+1]\pmod {13}\\
&\equiv [9(-1)^{4k}\cdot 3(-1)^{2k}+1]\pmod {13}\\
&\equiv (9+3+1)\pmod {13}\\
&\equiv 0\pmod {13}.\\
\end{align*}
Thus, $13\mid (10^{2p}-10^{p}+1)$.
Case #2: Suppose $p=6k+5$ for some $k\in\mathbb{Z}$.
Then $10^{2p}-10^{p}+1=10^{12k+10}-10^{6k+5}+1$.
Observe that
\begin{align*}
&10^{2p}-10^{p}+1\equiv [(-3)^{12k+10}-(-3)^{6k+5}+1]\pmod {13}\\
&\equiv [-3((-3)^{3})^{4k+3}\cdot (-9)((-3)^{3})^{2k+1}+1]\pmod {13}\\
&\equiv [-3(-1)^{4k+3}\cdot (-9)(-1)^{2k+1}+1]\pmod {13}\\
&\equiv (3+9+1)\pmod {13}\\
&\equiv 0\pmod {13}.\\
\end{align*}
Thus, $13\mid (10^{2p}-10^{p}+1)$.
Therefore, $13$ divides $10^{2p}-10^{p}+1$ for any prime $p>3$.

 

[fresh_42]

Homework Statement:: For any prime $p>3$, prove that $13$ divides $10^{2p}-10^{p}+1$.
Relevant Equations:: None.

Proof:

Let $p>3$ be any prime number.
Then $p>3$ is either of the form $6k+1$ or $6k+5$ for some $k\in\mathbb{Z}$.
Now we consider two cases.
Case #1: Suppose $p=6k+1$ for some $k\in\mathbb{Z}$.
Then $10^{2p}-10^{p}+1=10^{12k+2}-10^{6k+1}+1$.
Observe that
\begin{align*}
&10^{2p}-10^{p}+1\equiv [(-3)^{12k+2}-(-3)^{6k+1}+1]\pmod {13}\\
&\equiv [9((-3)^{3})^{4k}\cdot 3((-3)^{3})^{2k}+1]\pmod {13}\\
&\equiv [9(-1)^{4k}\cdot 3(-1)^{2k}+1]\pmod {13}\\
&\equiv (9+3+1)\pmod {13}\\
&\equiv 0\pmod {13}.\\
\end{align*}
Thus, $13\mid (10^{2p}-10^{p}+1)$.

Correct, except that your \cdot should be +.

Case #2: Suppose $p=6k+5$ for some $k\in\mathbb{Z}$.
Then $10^{2p}-10^{p}+1=10^{12k+10}-10^{6k+5}+1$.
Observe that
\begin{align*}
&10^{2p}-10^{p}+1\equiv [(-3)^{12k+10}-(-3)^{6k+5}+1]\pmod {13}\\
&\equiv [-3((-3)^{3})^{4k+3}\cdot (-9)((-3)^{3})^{2k+1}+1]\pmod {13}\\
&\equiv [-3(-1)^{4k+3}\cdot (-9)(-1)^{2k+1}+1]\pmod {13}\\
&\equiv (3+9+1)\pmod {13}\\
&\equiv 0\pmod {13}.\\
\end{align*}
Thus, $13\mid (10^{2p}-10^{p}+1)$.

Same as above, \cdot should be +, but otherwise correct.

Therefore, $13$ divides $10^{2p}-10^{p}+1$ for any prime $p>3$.