×

# Only for advanced number theorists

1.| Show that $$n^{4}$$ - $$20n^{2}$$ + 4 is composite when n is any integer.

2.| Prove that there are no prime numbers in the infinite sequence of integers 10001, 100010001, 1000100010001,...........

Note by Swapnil Rajawat
3 years, 3 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:

\begin{align} n^4-20n^2+4-16n^2 &= (n^2-2)^2-(4n)^2\\ &=(n^2+4n-2)(n^2-4n-2) \end{align}

Now, none of the above two brackets are equal to $$1$$ for an integer $$n$$.

Also, for integer values of $$n$$, both the brackets are also integers.

- 3 years, 3 months ago

For the second problem: If there are an even number of 1's, the number is divisible by $$10001 = 73 \cdot 137$$. If there are an odd number, say $$b$$, the number is divisible by $$(10^b-1)/9$$. The proof is left to the reader.

- 3 years, 3 months ago

1. $$n^4-4n^2+4-16n^2=(n-2)^2-(4n)^2$$ which is the difference of the squares

- 3 years, 3 months ago

(Hint For second question) The terms of the sequence can be written as 1 + $$10^{4}$$, 1+$$10^{4}$$+$$10^{8}$$,.......1+$$10^{4}$$.......+$$10^{4n}$$..... more generally then the sequence is 1 + $$x^{4}$$, 1+$$x^{4}$$+$$x^{8}$$,.......1+$$x^{4}$$.......+$$x^{4n}$$..... for an arbitrary integer x, x > 1 If n is odd, say n = 2m + 1.

- 3 years, 3 months ago