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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

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

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.

Pratik Shastri - 3 years, 3 months ago

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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 - 3 years, 3 months ago

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

Math Man - 3 years, 3 months ago

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(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 - 3 years, 3 months ago

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