Find all the natural numbers between $1 \leq a \leq 1000$ which fulfill $a^{11764} \equiv a^4\pmod{25725}$.
What I have tried so far was reducing $a$ and factoring the modulus to a smaller number related to $5$.
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5Note: $25725 = 3 * 5^2 * 7^3$. By the Chinese Remainder theorem, the result will hold if and only if $a^{11764} \equiv a^4$ holds modulo $3$, modulo $5^2$, and modulo $7^3$. – Ben Grossmann Dec 27 '21 at 21:37
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Are you aware of the Fermat-Euler theorem? Do you see how this result is useful here? – Ben Grossmann Dec 27 '21 at 21:40
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You can begin by putting $ b=a^4$. – hamam_Abdallah Dec 27 '21 at 21:41
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@BenGrossmann This was the clue i was looking for, it solves my question. Thank you very much sir! – Prime_Number Dec 27 '21 at 21:45
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1The simple answer is: all of them. – Thomas Andrews Dec 27 '21 at 21:52
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Actually, $a^{11763} \equiv a^3\pmod{25725}$ for all $a \in \mathbb Z$. See https://en.wikipedia.org/wiki/Carmichael_function#Exponential_cycle_length – lhf Dec 28 '21 at 00:13
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1Special case of the Theorem in the linked dupe. Here $,11760 = \phi(25725)\ \ $ – Bill Dubuque Dec 29 '21 at 13:17