Exercise 1.21

Proof. As $a\wedge b\mid a+b,{\mathrm{ord}}_p(a\wedge b)\le {\mathrm{ord}}_p(a+b)$, so $\min <!--\; FUNCTION\; APPLICATION\; -->({\mathrm{ord}}_p(a),{\mathrm{ord}}_p(b))\le {\mathrm{ord}}_p(a+b)$.

Suppose that ${\mathrm{ord}}_p(a)\ne {\mathrm{ord}}_p(b)$. The problem being symmetric in $a,b$, we may suppose $\alpha ={\mathrm{ord}}_p(a)<\beta ={\mathrm{ord}}_p(b)$. So there exist $q,r\in \mathbb{Z}$ such that

Then $a+b=p^{\alpha }(q+p^{\beta -\alpha }r)$, where $p\nmid q+p^{\beta -\alpha }r$ (as $p\mid p^{\beta -\alpha }$ and $p\nmid q$).

So ${\mathrm{ord}}_p(a+b)=\alpha =\min <!--\; FUNCTION\; APPLICATION\; -->({\mathrm{ord}}_p(a),{\mathrm{ord}}_p(b))$. □