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Prime factorization of a huge number

Hi!

Does anyone know how to solve this problem:

Find one prime factor of \(1+{ 2 }^{ 21 }+{ 4 }^{ 21 }\).

Thanks for any help in advance!

Note by Julian Yu
4 months ago

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Let \(x=2^3\). We get the expression as \(x^{14}+x^7+1\).

Note that \(\omega\) and \(\omega^2\) divides this expression where \(\omega,\omega^2\) are the complex cube roots of unity.

The expression which has its roots as the complex cube roots of unity is \(x^2+x+1\).

Therefore \(x^2+x+1\) divides \(x^{14}+x^7+1\) or \(2^6+2^3+1=73\) divides \(1+2^{21}+4^{21}\). Svatejas Shivakumar · 4 months ago

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