# A Strange Number Theory proof?

There exists positive integers $$n$$ such that $$2n + 1$$ and $$3n + 1$$ are perfect squares. Prove that $$n$$ is divisible by $$40$$.

Note by Sharky Kesa
2 years, 9 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

If $$n$$ is odd, then $$2n+1=2(2k+1)+1=4k+3$$. But $$3$$ is not a quadratic residue mod $$4$$.

If $$n\equiv \{2,4,6\}\pmod{8}$$, then $$3n+1\equiv\{7,5,3\}\not\equiv x^2\pmod{8}$$. Therefore $$8\mid n$$.

If $$n\equiv\{1,3\}\pmod{5}$$, then $$2n+1\equiv\{3,2\}\not\equiv x^2\pmod{5}$$.

If $$n\equiv\{2,4\}\pmod{5}$$, then $$3n+1\equiv\{2,3\}\not\equiv y^2\pmod{5}$$. So $$5\mid n$$.

- 2 years, 9 months ago

Actually $$n$$ is divisible by 40 as well. Try it. I've a solution, but I'm awaiting others to post theirs. Meanwhile, I'd say that the problem would have been much accurate had you asked us to prove that $$n$$ is divisible by 40 instead of 8.

- 2 years, 9 months ago

You could just post it directly to me on Slack.

- 2 years, 9 months ago

Bhai, can you come on Slack ?

- 2 years, 9 months ago

Sure. Give me a few moments.

- 2 years, 9 months ago

Of Course!

- 2 years, 9 months ago