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,...........

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

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.

Log in to reply

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.

Log in to reply

Log in to reply

(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.

Log in to reply