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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 .

Note by Chinmay Sangawadekar
10 months, 2 weeks 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 \(p_2^{p_1}\equiv p_2\pmod{n}\).

Prasun Biswas - 10 months, 2 weeks ago

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