Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

Proving that $p\mid (p+1776)$ if $p$ is a prime and $p(p+1776)$ is a perfect square

+4
−0

Problem: Suppose $p$ is a prime number and $p(p+1776)$ is a perfect square. Prove that $p\mid (p+1776)$.

From the assumption of the problem, $p(p+1776)=k^2$ for some positive integer $k$. This does not help much. Intuitively, one can write $p(p+1776)=p^2m^2$ for some integer $m$ due to the fundamental theorem of arithmetic: in order to have a complete square, the factor $p$ should appear twice. One can then easily conclude that $p\mid (p+1776)$ since $p+1776=pm^2$. But I don't find a formal way to show this intuition.

History
Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

0 comment threads

2 answers

You are accessing this answer with a direct link, so it's being shown above all other answers regardless of its score. You can return to the normal view.

+4
−0

You basically have it.

I'll give an overly detailed rendition of an informal proof below.

As you state, the assumption is that $p(p+1776) = k^2$ for some natural number $k$ with $p$ a prime. We want to show that $p\mid p + 1776$, or, equivalently, $p + 1776 = pm$ for some natural $m$.

Proof: Via the Fundamental Theorem of Arithmetic, $k$ has some unique prime factorization so $k = \prod_i {p_i}^{n_i}$ where all the $p_i$ are prime. $k^2$ is then clearly $\prod_i ({p_i}^{n_i})^2$. Since $p \mid p(p + 1776)$, we have $p \mid k^2$ so there's some $i$ such that $p = p_i$. This is because for a prime $p$, $p \mid rs$ if and only if $p \mid r$ or $p \mid s$, and $p \mid q$ for a prime $q$ if and only if $q = p$. In symbols we have $$p \mid \prod_i {p_i}^{2n_i} \iff \bigvee_i p \mid {p_i}^{2n_i} \iff \bigvee_i p \mid p_i \iff \exists i. p = p_i$$ where $\iff$ is "if and only if" and $\bigvee$ is a variadic disjunction. Choose $j$ such that $p = p_j$.

We thus have $p(p+1776) = k^2 = p^2 m$ where $m = p^{2n_i - 2} \prod_{i \neq j} {p_i}^{2n_i}$, so by cancelling $p$ from both sides we have $p + 1776 = p m$ which is exactly what it means for $p$ to divide $p + 1776$. $\square$

The Fundamental Theorem of Arithmetic is likely overkill for this problem, so it may be interesting to find a proof that doesn't rely on it.

History
Why does this post require moderator attention?
You might want to add some details to your flag.

1 comment thread

Thank you for your answer! I see from your proof that the crucial step is the fact that for prime $p$... (1 comment)
+4
−0

Inspired by Derek Elkins's excellent answer, 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 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 of the fundamental theorem of arithmetic.

History
Why does this post require moderator attention?
You might want to add some details to your flag.

0 comment threads

Sign up to answer this question »