Considering mod 19 may be useful

There are no solutions in integers... Consider $mod \ 19$. Then as $20 \equiv 1 \pmod {19}$ so $20^x \equiv 1 \pmod {19}$. Again, $13^2 \equiv -2 \pmod {19}...(i)$, and $2013 \equiv -1 \pmod {19}$. So $2013^z \equiv 1,-1 \pmod {19}$. Now note this forces $z$ ti be odd as $20^x+13^2 \equiv -1 \pmod {19}$ and taking any higher power of $2$ in $(i)$ will never again result in a $-2$ as remainder, so $y=2,z=2k+1$. Now notice that $2013^{2k+1} \equiv x \pmod {10}$ and $169 \equiv 9 \pmod {10}$ so $2013^{2k+1}-169 \equiv x-9 \pmod {10}$. So this forces $x$ to be $-1$, as $20^k \equiv 0 \pmod {10}$. So this basically implies that $2013^{2k+1} \equiv -1 \pmod {10}$, but also note that $2013 \equiv 3 \pmod {10} \implies 2013^2 \equiv -1 \pmod {10}$. And so, $\forall$ odd $l, 2013^{2l} \equiv -1 \pmod {10}$, but $2l$ is even, and $z$ must be odd, so this is a contradiction and no solutions exist. $\qquad \blacksquare$

by considering the equation mod 7 $$(-1)^x + (-1)^y = 4^z$$mod 7 now $4^z \in \{4,2,1\}$ and the only possible numbers we can make on the LHS are $\{-2,0,2\}$ hence we know that: $x \equiv y \equiv 0$ mod 2 and that $z \equiv 2$ mod 3. Now checking mod 3: $(-1)^x + 1^y = (-1)^x + 1 \equiv 0$ mod 3. So now we find that x is even, an obivous contradiction

Go mod 10 at first. You will get that y=z . Now apply zsigmondy theorem by transferring the terms with same exponents and conclude it with no solutions at all.

Why to memorize extra theorems?

Rizsgtp wrote: Go mod 10 at first. You will get that y=z . Now apply zsigmondy theorem by transferring the terms with same exponents and conclude it with no solutions at all. $$3^5 \equiv 3 \pmod{10}$$doesn't imply $5=1$?

khan.academy wrote: Rizsgtp wrote: Go mod 10 at first. You will get that y=z . Now apply zsigmondy theorem by transferring the terms with same exponents and conclude it with no solutions at all. $$3^5 \equiv 3 \pmod{10}$$doesn't imply $5=1$? No because we are only looking at the remainder mod 10. In fact phi(10)=4 so x^5 = x mod 10 whenever (x,10)=1

WeakMathemetician wrote: khan.academy wrote: Rizsgtp wrote: Go mod 10 at first. You will get that y=z . Now apply zsigmondy theorem by transferring the terms with same exponents and conclude it with no solutions at all. $$3^5 \equiv 3 \pmod{10}$$doesn't imply $5=1$? No because we are only looking at the remainder mod 10. In fact phi(10)=4 so x^5 = x mod 10 whenever (x,10)=1 Okay, I am confused on the part why $y=z$? Can you please clarify?

Okay, I am confused on the part why $y=z$? Can you please clarify?[/quote] Its not my solution i assume they meant y = z mod 10. Idk i think his solution may be flawed

Can $x,y,z$ be negative?

Here's my solution.

Take modulo $7$ to get that $(-1)^x + (-1)^y \equiv 4^z \mod 7$. Clearly, since $4^0 \equiv 1 \mod 7, 4^1 \equiv 4 \mod 7, 4^2 \equiv 2 \mod 7, 4^3 \equiv 1 \mod 7$, we see that whenever $n \equiv 0 \mod 3$, $4^n \equiv 1 \mod 7$, whenever $n \equiv 1 \mod 3, 4^n \equiv 4 \mod 7$, and whenever $n \equiv 2 \mod 3$, $4^n \equiv 2 \mod 7$. Back the game-board: $(-1)^x + (-1)^y$ can only take on the values $-2,0,2$, so by examining all possible values of $7^z$, we see that $(-1)^x + (-1)^y = 2$ is the only possibility. This happens only when $x$ and $y$ are both even. On the other hand, we can take modulo $3$ to get that $(-1)^x + 1^y \equiv 0 \mod 3$ which implies that $(-1)^x \equiv (-1) \mod 3$ which means that $x$ is odd. This directly contradicts what we showed beforehand, so there clearly aren’t any solutions.

Notice that: $\implies 20^x \equiv 6\pmod7\equiv-1\pmod7$ $\implies 13^y\equiv6\pmod7\equiv-1\pmod7$ $\implies 2013^z\equiv4\pmod7$ So now we can convert everything into what we found above. $\implies (-1)^x + (-1)^y \equiv 4^z$ We clearly see that: $\implies (-1)^x \equiv 0,1\pmod7$ $\implies (-1)^y\equiv0,1\pmod7$ (Well, this is true for any positive integer $q$ really...) So, we test all possible cases.. $\implies (-1)^x + (-1)^y \equiv -2,0,2\pmod7$ (Again, this is true for any positive integer $q$...) Recall the equation we found: $\implies (-1)^x + (-1)^y \equiv 4^z\pmod7$ Notice that: $\implies 4^z\equiv4,2,1\pmod7$ Notice that $(-2,0,2)\cap(4,2,1)=\vartheta$. This implies that there are no integer solutions.

We have $(-1)^x+1^y\equiv 0\mod 3$ so $x$ is odd. Also $(-2)^x+2^y=-2^x+2^y\equiv 0\mod 11\iff 2^x\equiv 2y\mod 11$ so $y\equiv x\mod 10$ so have same parity. Finally $3^y\equiv 3^z\mod 5$ so $y\equiv z\mod 4$ indeed $x,y,z$ are odd.

Zone wrote: Can $x,y,z$ be negative? If so, all of $x,y,z$ must be negative at the same time. Setting $ x=-a, y=-b, z=-c $ $ \frac{1}{20^a} + \frac{1}{13^b} = \frac{1}{2013^c} $ Multiplying both sides by denominators, you'll get that $c=0$. Then there doesn't exist any integer numbers $a, b$ , so $x, y, z$.