- #1

Math100

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- 220

- Homework Statement
- For ## n\geq 1 ##, use congruence theory to establish the following divisibility statement:

## 7\mid (5^{2n}+3\cdot 2^{5n-2}) ##.

- Relevant Equations
- None.

Proof:

Let ## n\geq 1 ## be a natural number.

Note that ## 5^{2}\equiv 4\pmod 7\implies (5^{2})^{n}\equiv 4^{n}\pmod {7} ##.

Now observe that \begin{align*} (3\cdot 2^{5n-2})&\equiv (3\cdot 2^{3}\cdot 2^{5n-5})\pmod {7}\\

&\equiv [3\cdot 2^{3}\cdot (2^{5})^{n-1}]\pmod {7}\\

&\equiv [24\cdot (2^{5})^{n-1}]\pmod {7}\\

&\equiv (3\cdot 4^{n-1})\pmod {7}\\

&\equiv (-4\cdot 4^{n-1})\pmod {7}\\

&\equiv -(4^{n})\pmod {7}.

\end{align*}

Thus ## 5^{2n}+3\cdot 2^{5n-2}\equiv (4^{n}-4^{n})\pmod 7\equiv 0\pmod 7 ##.

Therefore, ## 7\mid (5^{2n}+3\cdot 2^{5n-2}) ## for ## n\geq 1 ##.

Let ## n\geq 1 ## be a natural number.

Note that ## 5^{2}\equiv 4\pmod 7\implies (5^{2})^{n}\equiv 4^{n}\pmod {7} ##.

Now observe that \begin{align*} (3\cdot 2^{5n-2})&\equiv (3\cdot 2^{3}\cdot 2^{5n-5})\pmod {7}\\

&\equiv [3\cdot 2^{3}\cdot (2^{5})^{n-1}]\pmod {7}\\

&\equiv [24\cdot (2^{5})^{n-1}]\pmod {7}\\

&\equiv (3\cdot 4^{n-1})\pmod {7}\\

&\equiv (-4\cdot 4^{n-1})\pmod {7}\\

&\equiv -(4^{n})\pmod {7}.

\end{align*}

Thus ## 5^{2n}+3\cdot 2^{5n-2}\equiv (4^{n}-4^{n})\pmod 7\equiv 0\pmod 7 ##.

Therefore, ## 7\mid (5^{2n}+3\cdot 2^{5n-2}) ## for ## n\geq 1 ##.

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