$x^{41}\equiv y^{49}\equiv a \pmod p$ want to find $z$ st $z^{2009}\equiv a \pmod p$ if $p\mid a$, then let $z\equiv 0 \pmod p$ if $p\nmid a$, then let $z\equiv x^b y^c$ where $b,c\in\mathbb{Z}$ $z^{2009}\equiv x^{2009b} y^{2009c}\equiv a^{49b} a^{41c}\equiv a^{49b+41c}\equiv a \pmod p$ for last congruence, b,c exist because gcd(49,41)=1 We know the residue class of z modulo p. But we need $z^{2009} \geq a$ So assume $z\equiv d \pmod p$ to get $z$, pick one number $\equiv d \pmod p$ and $\geq a$ then $z^{2009} \geq a^{2009} \geq a$