How can I prove that expression using modular methods ?
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What have you tried? What are you stuck on? Please edit your question to include more information. – Malady Dec 25 '23 at 19:58
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I tried to simplify the expression and settled on $2^n(4 \cdot5^{2n}+ 15 \cdot 3^n2^n) \equiv 0 \pmod {19}$ – Шамиль Замалиев Dec 25 '23 at 20:05
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Rewrite as $\ a, 50^{n-1}! + b, 12^{n-1}! \equiv (a!+!b),12^{n-1}\equiv 0,$ by $,50\equiv 12,,a+b\equiv 0\pmod{!19},,$ by linked congruence rules. $\ \ $ – Bill Dubuque Dec 25 '23 at 20:28
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Yes I did it with $12^{n-1}$, thank you – Шамиль Замалиев Dec 25 '23 at 20:30