Prove that the integer ## 53^{103}+103^{53} ## is divisible by....

[Math100]

Homework Statement

Prove that the integer $53^{103}+103^{53}$ is divisible by $39$, and that $111^{333}+333^{111}$ is divisible by $7$.

Relevant Equations

None.

Proof:

First, we will prove that the integer $53^{103}+103^{53}$ is divisible by $39$.
Note that $53\equiv 14 \pmod {39}\implies 53^{2}\equiv 14^{2}\pmod {39}\equiv 196\pmod {39}\equiv 1\pmod {39}$.
Now observe that $103\equiv 25\pmod {39}\equiv -14\pmod {39}\implies 103^{2}\equiv 196\pmod {39}\equiv 1\pmod {39}$.
Thus $53^{103}+103^{53}\equiv (53^{2})^{51}\cdot 53+(103^{2})^{26}\cdot 103\equiv (1^{51}\cdot 53+1^{26}\cdot 103)\pmod {39}\equiv 156\pmod {39}\equiv 0\pmod {39}$.
Therefore, the integer $53^{103}+103^{53}$ is divisible by $39$.
Next, we will prove that the integer $111^{333}+333^{111}$ is divisible by $7$.
Note that $111\equiv 6\pmod 7\equiv (-1)\pmod 7\implies 111^{333}\equiv (-1)^{333}\pmod 7\equiv (-1)\pmod 7$.
Now observe that $333=3\cdot 111\equiv 3\cdot (-1)\pmod 7\equiv -3\pmod 7\equiv 4\pmod 7\implies 333^{3}\equiv 4^{3}\pmod 7\equiv 1\pmod 7\implies 333^{111}\equiv (333^{3})^{37}\equiv 1^{37}\pmod 7\equiv 1\pmod 7$.
Thus $111^{333}+333^{111}\equiv (-1+1)\pmod 7\equiv 0\pmod 7$.
Therefore, the integer $111^{333}+333^{111}$ is divisible by $7$.

 

[fresh_42]

Homework Statement:: Prove that the integer $53^{103}+103^{53}$ is divisible by $39$, and that $111^{333}+333^{111}$ is divisible by $7$.
Relevant Equations:: None.

Proof:

First, we will prove that the integer $53^{103}+103^{53}$ is divisible by $39$.
Note that $53\equiv 14 \pmod {39}\implies 53^{2}\equiv 14^{2}\pmod {39}\equiv 196\pmod {39}\equiv 1\pmod {39}$.
Now observe that $103\equiv 25\pmod {39}\equiv -14\pmod {39}\implies 103^{2}\equiv 196\pmod {39}\equiv 1\pmod {39}$.
Thus $53^{103}+103^{53}\equiv (53^{2})^{51}\cdot 53+(103^{2})^{26}\cdot 103\equiv (1^{51}\cdot 53+1^{26}\cdot 103)\pmod {39}\equiv 156\pmod {39}\equiv 0\pmod {39}$.
Therefore, the integer $53^{103}+103^{53}$ is divisible by $39$.
Next, we will prove that the integer $111^{333}+333^{111}$ is divisible by $7$.
Note that $111\equiv 6\pmod 7\equiv (-1)\pmod 7\implies 111^{333}\equiv (-1)^{333}\pmod 7\equiv (-1)\pmod 7$.
Now observe that $333=3\cdot 111\equiv 3\cdot (-1)\pmod 7\equiv -3\pmod 7\equiv 4\pmod 7\implies 333^{3}\equiv 4^{3}\pmod 7\equiv 1\pmod 7\implies 333^{111}\equiv (333^{3})^{37}\equiv 1^{37}\pmod 7\equiv 1\pmod 7$.
Thus $111^{333}+333^{111}\equiv (-1+1)\pmod 7\equiv 0\pmod 7$.
Therefore, the integer $111^{333}+333^{111}$ is divisible by $7$.

Correct. If you want to avoid too long lines, then you can use the following structure:

\begin{align*}
53^{103}+103^{53}&\equiv (53^{2})^{51}\cdot 53+(103^{2})^{26}\cdot 103\pmod {39}\\
&\equiv (1^{51}\cdot 53+1^{26}\cdot 103)\pmod {39}\\
&\equiv 156\pmod {39}\\
&\equiv 0\pmod {39}
\end{align*}

which results in

\begin{align*}
53^{103}+103^{53}&\equiv (53^{2})^{51}\cdot 53+(103^{2})^{26}\cdot 103\pmod {39} \\
&\equiv (1^{51}\cdot 53+1^{26}\cdot 103)\pmod {39}\\
&\equiv 156\pmod {39}\\
&\equiv 0\pmod {39}
\end{align*}