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# Prime on prime, gives base prime

Let $n=p_{1} × p_{2}$ where $$n$$ belongs to the set of natural numbers and $$p_{1}$$, $$p_{2}$$ are prime factors of $$n$$ , such that $$p_{1}<p_{2}$$ .

Then $p_{1}^{p_{2}} \equiv p_{1} \mod n$ holds true .

8 months, 1 week ago

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This is trivial. We have the obvious congruence $$p_1^{p_2}\equiv 0\equiv p_1\pmod{p_1}$$ since $$p_1\mid p_1^{p_2}$$.

Since $$p_1,p_2$$ are distinct primes, we have $$\gcd(p_1,p_2)=1$$ and hence, by Fermat's Little Theorem, we have $$p_1^{p_2}\equiv p_1\pmod{p_2}$$.

From the above results, we see that $$p_1^{p_2}\equiv p_1\pmod{p_1,p_2}$$. Since $$\gcd(p_1,p_2)=1$$, we have $$p_1^{p_2}\equiv p_1\pmod{p_1p_2}$$, i.e., $$p_1^{p_2}\equiv p_1\pmod{n}$$.

From the above, notice that since we just use the fact that $$p_1,p_2$$ are distinct and not that $$p_1\lt p_2$$, we also have $$p_2^{p_1}\equiv p_2\pmod{n}$$. · 8 months, 1 week ago