Given the sequence $(a_n)_{n\geq 1}$ with $a_n = \frac{{2 \cdot 2^{2^{n}} + 1}}{3}$, prove that $3^n \mid a_{3^n}$ for every $n \geq 1$.
I thought about using LTE 2 times for the numerator but I got stuck.
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math.enthusiast9
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I wonder how you apply LTE, considering the 2 multiplying the 4. – D S Mar 18 '24 at 17:57
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Oh, you've edited the question and removed the three. – D S Mar 18 '24 at 18:40
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I feel I must warn you that your question is lacking context and will be closed soon. – D S Mar 18 '24 at 18:41
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You are continuously changing the problem statement. What is the question exactly? – D S Mar 18 '24 at 18:55
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@DS -- the problem as stated earlier was incorrect. OP has resolved the problem in chat. – Sahaj Mar 18 '24 at 19:02
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where exactly? @Sahaj – D S Mar 18 '24 at 19:07
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1To the OP: This is a popular lemma in disguise. You will need $a \equiv b \pmod {m^n} \implies a^m \equiv b^m \pmod{m^{n+1}}$ along with LTE. – D S Mar 18 '24 at 19:11
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Upon more thinking, applying LTE twice should also have worked. If you show how you failed while using LTE, the question may be reopened. – D S Mar 18 '24 at 19:25