For ## n\geq 1 ##, use congruence theory to establish....

[Math100]

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 $n\geq 1$.

 

[fresh_42]

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 $n\geq 1$.

Correct, up to typos in the last line of the align environment which I corrected. You had { ... ) which does not match and $-4\cdot 4^{n-1}=(-4^{n-1})$ which was wrong.

 

[Math100]

Thank you!

 

[Mark44]

@Math 100, you have started several threads with very generic and uninformative titles such as "For $n\geq 1$, use congruence theory to establish?". For future threads, please provide titles that indicate what it is you're trying to prove. For this thread, a better title would be "Prove that $7\mid (5^{2n}+3\cdot 2^{5n-2})$".
It's not necessary to give all of the details in the thread, such as $n \ge 1$.