×

# Can u prove it?

Prove that n^4+4 is a composite number for n>1?

Note by Naitik Sanghavi
2 years, 8 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:

Note that $$n^{4} + 4 = (n^{4} + 4n^{2} + 4) - 4n^{2} = (n^{2} + 2)^{2} - (2n)^{2} = ((n^{2} + 2) - 2n)((n^{2} + 2) + 2n).$$

For $$n \gt 1$$ both of these last two bracketed terms are $$\gt 1,$$ thus proving that $$n^{4} + 4$$ is composite for $$n \gt 1.$$

- 2 years, 8 months ago

Perfect!!!..

- 2 years, 8 months ago

$$a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)$$

$$(a,b)=(n,1)$$ gives $$n^4+4=(n^2+2+2n)(n^2+2-2n)$$,

which is composite for $$n\ge 2$$, since $$n^2+2-2n\ge 2$$.

- 2 years, 8 months ago