Show that the integer $53^{103} + 103^{53}$ is divisible by $39$.

Example: Show that the integer $53^{103} + 103^{53}$ is divisible by $39$.

 Answer: Firstly, we will find the remainder when $53^{103} $ and $ 103^{53}$ is divided by $39$. Then with help of property $4.$ of Theorem, we show that  the integer $53^{103} + 103^{53}$ is divisible by $39$.

We have $53 \equiv 14 ~(mod ~39)$. By property $6.$ of Theorem, we have $53^2 \equiv 14^2 \equiv 1 ~(mod ~39)$. We can write $103$ as $103 = 51 \cdot 2 + 1$. Therefore $53^{103} \equiv (53^{2})^{51} \cdot 53 \equiv 1^{51} \cdot 53 \equiv  14 ~(mod ~39)$, i.e. $53^{103} \equiv  14 ~(mod ~39)$. This gives that $14$ is a remainder when $53^{103}$ is divided by $39$.

We have $103 \equiv 25 ~(mod ~39)$. But $25 \equiv -14 ~ (mod ~ 39)$. Therefore $103 \equiv -14 ~(mod ~39)$. By property $6.$ of Theorem, we have $103^2 \equiv (- 14)^2 \equiv 1 ~(mod ~39)$.  We can write $53$ as $53 = 26 \cdot 2 + 1$. Therefore $103^{53} \equiv (103^{2})^{26} \cdot 103 \equiv 1^{26} \cdot 103 \equiv  25 ~(mod ~39)$, i.e. $103^{53} \equiv  25 ~(mod ~39)$. This gives that $25$ is a remainder when $103^{53}$ is divided by $39$.

By property $4.$ of Theorem, we have $53^{103} + 103^{53} \equiv 14 + 25 \equiv 0 ~(mod ~ 39)$. This show that  $0$ is a remainder when the integer $53^{103} + 103^{53}$ is divided by $39$, i.e. the integer $53^{103} + 103^{53}$ is divisible by $39$.

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