×

# Number Theory #4

Let $$n$$ be a positive integer such that $$2n+1$$ and $$3n+1$$ are both squares. Prove that $$40|n$$.

Note by Victor Loh
3 years, 6 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:

It suffices to prove $$5\mid n,8\mid n$$.

Given that $$2n+1=x^2, 3n+1=y^2$$, If we add the two equations we get: $$5n+2=x^2+y^2\Rightarrow x^2+y^2\equiv 2\pmod 5$$, since the quadratic residues of mod $$5$$ are$$0,1,4$$, we can check that only $$x^2\equiv y^2\equiv 1$$ works, therefore $$2n\equiv 1-1\equiv 0\pmod 5\Rightarrow 5\mid n$$

The same analysis can be done for $$8$$, but we will multiply the second equation by $$2$$ first and then add: $$8n+3=x^2+2y^2\Rightarrow x^2+2y^2\equiv 3\pmod 8$$, since the quadratic residues of mod $$8$$ are $$0,1,4$$, we can check that only $$x^2\equiv y^2\equiv 1$$ works again, which means $$3n\equiv 1-1\equiv 0\pmod 8\Rightarrow 8\mod n$$ and we are done.

- 3 years, 6 months ago

Great!

- 3 years, 6 months ago