Balkan Mathematical Olympiad

Consider equation modulo $13$. We have $12^x+13^y-14^z\equiv (-1{)}^x+0-1(\mathrm{mod}13)$, and $13\nmid 2013$, so $x$ must be odd. Also, $2013^t\equiv -2(\mathrm{mod}13)$, which implies $t=12t_1+1$, $t_1\in {\mathbb{N}}_0$.

Consider equation modulo $3$. We have $12^x+13^y-14^z\equiv 1-(-1{)}^z(\mathrm{mod}3)$, so $2\mid z$. Now, if we consider equation modulo $7$ we get (by FLT) $2013^t=2013^{12t_1+1}\equiv 4(\mathrm{mod}7)$, so $4\equiv 12^x+13^y-14^z\equiv 5^x+(-1{)}^y+0(\mathrm{mod}13)$. Therefore, we need to examine the following two cases:

First case: $12^x\equiv 3(\mathrm{mod}7)$ and $13^y\equiv 1(\mathrm{mod}7)$, i.e. $x\equiv 5(\mathrm{mod}6)$ and $2\mid y$. Then $13^y\equiv 1(\mathrm{mod}8)$, $x\ge 5$, so $8\mid 12^x$, and $2013^t\equiv 5(\mathrm{mod}8)$, so $8\nmid 14^z$. Since $z$ is even this implies $z=2$. Also, $16\mid 12^x$, so $13^y\equiv 2013^t+196\equiv 13+4\equiv 1(\mathrm{mod}16)$ and it follows that $4\mid y$. Then $13^y\equiv 1(\mathrm{mod}5)$, $2013^t\equiv 3(\mathrm{mod}5)$, and $12^x\equiv 3-1+1=3(\mathrm{mod}5)$, so $x=4x_1+3$, $x_1\in {\mathbb{N}}_0$. Now $13^y\equiv 1(\mathrm{mod}17)$, since $13^2\equiv -1(\mathrm{mod}17)$ and $4\mid y$. Considering modulo $17$, equation becomes $12^x+1-9\equiv 7^t(\mathrm{mod}17)$, and it is enough to check residues of $12^3,12^7,12^{11},12^{15}$, and $7^1,7^5,7^9,7^{13}$. Both of them give the set of residues $\{6,7,10,11\}$. Since there are no two number in this set with difference $8$ or $9$ modulo $17$, equation does not have any solutions in this case.

Second case: $12^x\equiv 5(\mathrm{mod}7)$ and $13^y\equiv -1(\mathrm{mod}7)$, i.e. $x\equiv 1(\mathrm{mod}6)$ and $2\nmid y$. Then $13^y\equiv 5(\mathrm{mod}8)$, so $8\mid 12^x-14^z$. Now, $x=1$ and $z=2$, or $x>1$ and $z>2$ (i.e. $z\ge 4$). If $x>1$ and $z\ge 4$, then $16\mid 2013^t-13^y$. Since $2013^t\equiv 13(\mathrm{mod}16)$, we have that $13^y\equiv 13(\mathrm{mod}16)$, and therefore $y=4y_1+1$, $y_1\in {\mathbb{N}}_0$. Now, considering equation modulo $5$ we obtain $13^y\equiv 2013^t\equiv 3(\mathrm{mod}5)$, and $14^z\equiv 1(\mathrm{mod}5)$, which implies $12^x\equiv 1(\mathrm{mod}5)$. This is not possible since $x$ is odd. So, $x=1$, $z=2$, and the equation is equivalent to $13^y=2013^t+184$. Suppose $t>1$. Then $t\ge 13$, so $27\mid 2013^t$. Therefore $27\mid 13^y-184$, which implies $13^y\equiv 22(\mathrm{mod}27)$. Since $3\mid 13-1$, we have $27\mid 13^9-1$. Now, it is easy to check that $y\equiv 4(\mathrm{mod}9)$. On the other hand, by FLT, we have $13^9\equiv \pm 1(\mathrm{mod}19)$. Then $13^y=13^{9y_2+4}\equiv \pm 13^4\equiv \pm 4(\mathrm{mod}19)$. Also $2013\equiv -1(\mathrm{mod}19)$, so $2013^t\equiv -1(\mathrm{mod}19)$ which is a contradiction.

So $t=1$, which leads to the only solution of the equation $(x,y,z,t)=(1,3,2,1)$.