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

#NumberTheory #Numbers #ProblemSolving #Problems

Easy Math Editor

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Math	Appears 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}`	$\boxed{123}$

Comments

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

math man - 6 years, 9 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.

Patrick Corn - 6 years, 9 months ago

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

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.

Pratik Shastri - 6 years, 9 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.

swapnil rajawat - 6 years, 9 months ago

Math	Appears 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}`	$\boxed{123}$