Školjka

$2009$ daje $2$ mod $3$, pa onda $2^n$ daje $1$ mod $3$, $x=2a$ $\forall x\ge 3, 8|3^y5^z-2009$, kako $2009$ daje $1$, $3^y5^z$ mora dati $1$ mod $8$. Ako je $z$ paran, $3^y$ mora dati $1$ mod $8$, pa je onda i $y$ paran. Ako je $z$ neparan, $5^z3^y$ je kongruentno $5\cdot 3^y$ što daje ostatke $5,7$ mod $8$, pa to nije moguce. Dakle jedina moguća kombinacija je $(x/2,y/2,z/2)=(a,b,c), a,b,c,\mathbb\in{N}$ $(3^{y/2}5^{z/2}-2^{x/2})(3^{y/2}5^{z/2}+2^{x/2})=2009$ $3^{y/2}5^{z/2}-2^{x/2}=41,3^{y/2}5^{z/2}+2^{x/2}=49$ $3^{y/2}5^{z/2}=45, (x,y,z)=(4,4,2)$ $3^{y/2}5^{z/2}-2^{x/2}=1,3^{y/2}5^{z/2}+2^{x/2}=2009$ $3^{y/2}5^{z/2}=67\cdot 3 \cdot 5$ nema rjesenja $3^{y/2}5^{z/2}-2^{x/2}=7,3^{y/2}5^{z/2}+2^{x/2}=287$ $3^{y/2}5^{z/2}=49\cdot 3$, nema rješenja Jedino rješenje je $(x,y,z)=(4,4,2)$