An alternative (maybe easier) way to prove the last corollary could be to use q^(p-1) = 1 mod p and multiply the WHOLE equation by q to get q^p = q mod (pq) and then similarly p^q = p mod (pq) and then add.
EDIT: Sanket was right and I didn't understand. I'm leaving this comment up to preserve the conversation. That is a great observation, though I'm not sure if you can go directly from q^p ≡ q mod p to q^p ≡ q mod pq, at least without proving it with the lemma first. You can say that since q^p ≡ q mod p, then q^p - q ≡ 0 mod p and p | q^p - q. But q^p - q ≡ 0 mod q since q^p and q are both multiples of q, and so q | q^p - q as well. Therefore p*q | q^p - q by the previous lemma, and so q^p - q ≡ 0 mod pq and q^p ≡ q mod pq (which is what you said in your comment). In a similar way, we can say that p^q ≡ p mod pq. Therefore p^q + q^p ≡ p + q mod pq.
@@PunmasterSTP I think you have a typo, I went from q^(p-1) = 1 mod p (Fermat's little) to q^p = q mod pq. You can multiply modular equations with the same constant (including the base). You can also quickly verify it by seeing that p divides a - b then pm divides am - bm .
@@SanketAlekar Wow, you showed me the light! I either forgot or never thought that a ≡ b mod n implies that ca ≡ cb mod cn in addition to ca ≡ cb mod n. Your way seems far more elegant. I hope Michael Penn could weight in at some point, or perhaps he addresses this in another video and I just have to keep watching...
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Professor M. Penn, this is another theoretical lecture on Corollaries to Fermat's Little Theorem , however it's easy to follow.
great video ,thank you sir
this video was very helpful and easy to follow. I am currently in number theory struggling but this just cleared up a ton of confusion. Thanks
An alternative (maybe easier) way to prove the last corollary could be to use q^(p-1) = 1 mod p and multiply the WHOLE equation by q to get q^p = q mod (pq) and then similarly p^q = p mod (pq) and then add.
EDIT: Sanket was right and I didn't understand. I'm leaving this comment up to preserve the conversation.
That is a great observation, though I'm not sure if you can go directly from q^p ≡ q mod p to q^p ≡ q mod pq, at least without proving it with the lemma first.
You can say that since q^p ≡ q mod p, then q^p - q ≡ 0 mod p and p | q^p - q. But q^p - q ≡ 0 mod q since q^p and q are both multiples of q, and so q | q^p - q as well. Therefore p*q | q^p - q by the previous lemma, and so q^p - q ≡ 0 mod pq and q^p ≡ q mod pq (which is what you said in your comment). In a similar way, we can say that p^q ≡ p mod pq. Therefore p^q + q^p ≡ p + q mod pq.
@@PunmasterSTP I think you have a typo, I went from q^(p-1) = 1 mod p (Fermat's little) to q^p = q mod pq. You can multiply modular equations with the same constant (including the base). You can also quickly verify it by seeing that p divides a - b then pm divides am - bm .
@@SanketAlekar Wow, you showed me the light! I either forgot or never thought that a ≡ b mod n implies that ca ≡ cb mod cn in addition to ca ≡ cb mod n. Your way seems far more elegant. I hope Michael Penn could weight in at some point, or perhaps he addresses this in another video and I just have to keep watching...
Great work
FLT? More like FTL, because with these theorems we can solve number theory problems at warp speed!
😫🤍(: