Post History
#2: Post edited
by
Snoopy
·
2022-09-15T19:21:22Z (over 2 years ago)
- Inspired by Derek Elkins's excellent [answer](https://math.codidact.com/posts/287002/287003#answer-287003), I have the following proof.
By the assumption, we have $p(p+1776)=k^2$ for some integer $k$ and thus $p\mid k^2$. Then, [Euclid's lemma](https://en.wikipedia.org/wiki/Euclid%27s_lemma) implies that $p\mid k$. It follows that $k=mp$ for some integer $m$ and thus $(mp)^2=k^2=p(p+1776)$, which implies by cancellation that $m^2p=p+1776$. The proof is now complete.
- Inspired by Derek Elkins's excellent [answer](https://math.codidact.com/posts/287002/287003#answer-287003), I have the following proof.
- By the assumption, we have $p(p+1776)=k^2$ for some integer $k$ and thus $p\mid k^2$. Then, [Euclid's lemma](https://en.wikipedia.org/wiki/Euclid%27s_lemma) implies that $p\mid k$. It follows that $k=mp$ for some integer $m$ and thus $(mp)^2=k^2=p(p+1776)$, which implies by cancellation that $m^2p=p+1776$. The proof is now complete.
- ---
- Note: Euclid's lemma is also used in the *[proof](https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic#Proof)* of the fundamental theorem of arithmetic.
#1: Initial revision
by
Snoopy
·
2022-09-15T19:16:44Z (over 2 years ago)
Inspired by Derek Elkins's excellent [answer](https://math.codidact.com/posts/287002/287003#answer-287003), I have the following proof. By the assumption, we have $p(p+1776)=k^2$ for some integer $k$ and thus $p\mid k^2$. Then, [Euclid's lemma](https://en.wikipedia.org/wiki/Euclid%27s_lemma) implies that $p\mid k$. It follows that $k=mp$ for some integer $m$ and thus $(mp)^2=k^2=p(p+1776)$, which implies by cancellation that $m^2p=p+1776$. The proof is now complete.