11 June 2018

Two

\(2\) is the only even prime number.

The Fermat's last theorem states that there are no positive integers \(x\), \(y\) and \(z\) such that \(x^n+y^n=z^n\) for \(n>2\).

\(2+2=2\cdot2=2^2\)

\(2=\left(1+i\right)\left(1-i\right)\)

\(n^2\pm n\) is always divisible by \(2\). Why?

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