= Mahler-measure norm argument at measure two
Let an irreducible $f\in\mathbb Z[X]$ have $M(f)=2$ and a root outside the closed unit disk. Its leading coefficient must have absolute value one. The product $B$ of all roots outside the unit disk is then an algebraic integer with $|B|=2$; conjugate pairs make $B$ real, so $B=\pm2$. The quotient of the norm of any root by $B$ is both rational and an algebraic integer, and its absolute value is at most one. It is therefore $\pm1$, so every root has absolute norm two.
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