I found this a very beautiful problem. Thanks @dangerousliri!! Here is my solution. Taking mod $5$, we find that $0-1\equiv (-1)^z \pmod 5$. Hence, $z$ must be odd. Now, we show that $y=1$. Suppose $y\ge 2$. We have $45^x \equiv 5^x \equiv 5 \pmod{20}$ and $6^y\equiv 16 \pmod{20}$, so $2019^z \equiv 1 \pmod{20}$ which means $z$ is even, contradiction. Hence, $y=1$. The equation becomes $$45^x=2019^z+6$$Now we can conclude by taking $v_3$. Note that $v_3(2019^z+6)$ is at most 2. Also, $v_3(45^x)$ is even. Hence, $v_3(2019^z+6)=2$ It is $2$ only when $z=1$, which gives the solution $(2,1,1)$. EDIT: Alternatively, after getting $y=1$, mod 9 will also solve the problem immediately. (Is this really a N5? Maybe proposer didnt see these solutions?)

Hi, what about $45^x-5^y=2020^z$?

gghx wrote: Hi, what about $45^x-5^y=2020^z$? I thought about this also long time ago and I think I did solve it. We just have to consider $v_{5}(n)$ to solve it. First case when $x\neq y$ and then when $x=y$ using Zsigmondy should finish the problem when $x\neq y$ and some factorization would finish when $x=y$.

This is misplaced imho. Taking $\pmod{5}$ yields, $$-1 \equiv (-1)^z \pmod{5}$$ So $z$ is odd. Taking $\pmod{4}$ yields, $$1-2^y \equiv (-1)^{z} \equiv -1 \pmod{4}$$ So $y = 1$. Since $v_3(6)=1$, by $\pmod{9}$, either $x$ or $z$ is $1$. By visual observation, the only solution is $(x,y,z)=(2,1,1)$.

Steve12345 wrote: Find all positive integers $x, y, z$ such that $45^x-6^y=2019^z$ Proposed by Dorlir Ahmeti, Albania Use mod 5: $$-1 \equiv (-1)^z \pmod 5 \implies z \; \text{odd}$$Now use $v_3(n)$ to get (obviously contradiction if $\min\{v_3(45^x), v_3(6^y)\}=2x$): $$\min\{v_3(45^x), v_3(6^y)\}=z \implies y=z$$Now use mod 4: $$1+2^y \equiv -1 \pmod 4 \implies y=1$$So we know that $y=z=1$ so pluggin on the original diophane we get: $$45^x=2025 \implies x=2$$Thus the only solution is $(2,1,1)$ Thus we are done

First, If $x=2$ then, $2025-6^y=2019^z$. Then, the solution is $(x,y,z)=(2,1,1)$. For $x\ge3$, we divide it into 3 cases: Case 1: If $2x<y$ then, the equation becomes $3^{2x}(5^x-2^y3^{y-2x})=3^{z}673^{z}$ and $2x=z$ which implies $5^x-2^y3^{y-2x}=673^{2x}$. But, $5^x<673^{2x}$. So, there are no solution for this case. Case 2: If $2x=y$ then, the equation becomes $3^{2x}(5^x-2^{2x})=3^z673^z$ again, $2x=z$ and $5^x-2^{2x}<673^{2x}$. No solution for this one either. Case 3: If $2x>y$ then, the equation becomes $3^y(3^{2x-y}5^x-2^y)=3^z673^z$ and $y=z$ so that $3^{2x-y}5^x-2^y=673^y$. Taking modulo 5 yields, $-2^y \equiv (-2)^y (\mod 5)$ which leads to $y$ is odd. Since $y$ is odd then, $2x-y$ is odd so, $3^{2x-y}\equiv 3 (\mod 8)$ Now, if $y\ge3$ then, taking modulo 8 gives $3^{2x-y}5^x-2^y\equiv 3(5^x)\equiv 7 or 3 (\mod 8)$. But, $673^y \equiv 1 (\mod 8)$. So, there are no solution for $y\ge3$. If $y=2$ then, $45^x-36=2019^z$. Since $x\ge3$ then, $z=2$ but, this is not a solution. If $y=1$ then, $45^x-6=2019^z$. Then, $z=1$ but it is already written above. Hence, the only triplet satisfying the problem is $(2,1,1)$

Let $E_n$ be the equation$\pmod n$. $E_5\Rightarrow-1\equiv(-1)^z$, so $z$ is odd. $E_4\Rightarrow2^y\equiv2$, so we need $y=1$. $E_9\Rightarrow3^z\equiv3$, so we similarly have $z=1$. Then $45^x=2025$, so $x=2$. The solution is $\boxed{(x,y,z)=(2,1,1)}$.

We have $(x,y,z)=(2,1,1)$ if $y\geq2 \rightarrow 6^y \equiv 0 (mod 4)$ so $45^x - 6^y \equiv 1(mod 4) \rightarrow z $ is even we have $ 45^x \equiv 0 (mod 5)$ so $45^x-6^y \equiv -1 (mod 5) \rightarrow z$ is odd (conflict)

I forgot I had an AoPS Account Case 1: $2x > y$ We have $v_3(45^x - 6^y) = \min(2x, y) = y$, and $v_3(2019^z) = z$, so $y = z$. We have: $$3^{2x}\cdot5^x = 3^y\cdot2^y + 3^y \cdot 673^y = 3^y(2^y + 673^y) \implies 3^{2x - y} \cdot 5^x = 2^y + 673^y$$Since $2x > y$, we have $3 \mid 2^y + 673^y \implies 3 \mid 2^y + 1$, so $y$ is odd. Observe that $v_5(2^y + 673^y) = 2 + v_5(y)$ by LTE, so $2 + v_5(y) = x \implies y \geq 5^{x - 2}$. Notice that $5^{x - 2} > 2x$ for $x \geq 4$ (a simple induction suffices ), so $x = 1, 2, 3$, and taking cases for $y = 1, 3, 5$ gives $(x, y) = (2, 1)$ as the only solution. Case 2: $2x < y$ This means that $2x = z$ by using the same argument above. Notice that the equation rearranges to $45^x - 2019^z = 6^y$, and $x > z$ (because of size). However $2x = z \implies z > 2x$, a contradiction. Case 3: $2x = y$ We have $3^{2x}\cdot5^x - 3^{2x}\cdot2^{2x} = 3^{2x}(5^x - 4^x) = 3^{z}\cdot 673^{z}$. Since $3 \nmid 673$, we have $3^{2x} \mid 3^z \implies z \geq 2x$, a contradiction yet again. Therefore, our only solutions are $(x, y, z) = (2, 1, 1)$.

Too easy!