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# 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
2 years, 6 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. · 2 years, 6 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. · 2 years, 6 months ago

1. $$n^4-4n^2+4-16n^2=(n-2)^2-(4n)^2$$ which is the difference of the squares
· 2 years, 6 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. · 2 years, 6 months ago