Nivynum wrote: Find all all positive integers x,y,and z satisfying the equation $x^3=3^y7^z+8$ We get $x^3\equiv 2\pmod 3\Rightarrow x\equiv 2\pmod 3$. So we get: $v_3(x^3-2^3)=v_3(x-2)+v_3(3)=y\Rightarrow 0< v_3(x-2)=y-1$ $\Rightarrow \exists k\in \mathbb{Z}:\gcd(k,3)=1, x=3^{y-1}k+2$ $\Rightarrow 3^{3y-3}k^3+2\cdot3^{2y-2}k^2+4\cdot3^{y-1}k=3^y7^z\Rightarrow 3|4k$ It's contradiction. So this equation doesn't have any solution.

Dadgarnia wrote: So this equation doesn't have any solution. Then how about $11^3 = 3^3 7^2 + 8$?

Is this the unique solution?

mavropnevma wrote: Then how about $11^3 = 3^3 7^2 + 8$? We can show $3(3^{y-2}+1)^2=7^z-1$ and $y\equiv 1\pmod 2, z\equiv 0\pmod 2$. Can you solve this new equation?

If we write the equation in another way. $(x-2)(x^2+2x+4)=3^y7^z$ By using $x$ is odd it is easy to show that $gcd(x-2,x^2+2x+4)=3$ or $gcd(x-2,x^2+2x+4)=1$ From the equation above we can say $x-2=3^a7^b , x^2+2x+4=3^c7^d$ Since $x^2+2x+4=(x+1)^2+3$ it can't be divisible by $9$ so $c$ must be $1$ or $0$.Hence $d \ge 1$.So $b=0,d=z$ $1$)If $c=1$ $c=1 \Longrightarrow a=y-1 \Longrightarrow 3^{2y-3}+2\cdot3^{y-1}+4=7^z$ If we examine the equation in modulo $4$ we can see $z$ must be even.Let's write $z=2k$ in the equation. Hence $3^{y-1}(3^{y-2}+2)=(7^k-2)(7^ +2) $ Since $7^k \equiv 1 \pmod 3$ $7^k-2$ can't divisible by 3.Hence $3^{y-1}|7^k+2$. Hence $7^k+2 \ge 3^{y-1}$ and $7^k-2 \le 3^{y-2}+2 \Longrightarrow 3^{y-2}+6 \ge 3^{y-1} $ And the only solution is $y=3$. $2$)If $c=0$ $c=0 \Longrightarrow x-2=3^y , x^2+2x+4=7^z$ Then we have $3^{2y}+6\cdot3^y+12=7^z$ And this is easy to see that it has no solution in positive integers. So the only solution is $(11,3,2)$ $\Box$

Comment: I find this question easy for 2. It isnt necessary to use any special method. Too ordinary. And i also think there is much convenient problems in JBMO TST.

Considering modulo 3 one finds that $ x=3k + 2$, hence \[ y = val_3(x^3 - 8) = val_3 (x-2) + 1 \] which means \[ \underset{3^{y-1}}{(x-2)}\underset{3}{(x^2 + 2x + 4)}. \] Now let us consider valuation according to 7: Case I: If $ 7 | (x-2)$ then \[ z = val_7(x^3 - 8) = val_7(x-2) \] which means \[ \underset{7^{y}}{(x-2)}\underset{7^0}{(x^2 + 2x + 4)} \] combined with valuation according to 3 gives $ x-2 = 3^{y-1}7^z$ and $ x^2 + 2x + 4 = 3$ which gives no solution. Case II: If $ 7 \not | (x-2)$ then according to valuation 3 we had $ x-2 = 3^{y-1}$ and $ x^2 + 2x + 4 = 3\cdot 7^z.$ Plugging first one into the last equation and after cancellation we get \[ 3^{2y-3} + 2\cdot 3^{y-1} + 4 = 7^z \] also taking $ z=2w$ (considering $ \mod 4$ one finds that $ z$ is even) last equation can be rewritten as \[ 3^{y-1}(3^{y-4} + 2) = (7^w-2)(7^w + 2).\] Initial values of $ y = 1,2,3,4$ yield the only solution $ \boldsymbol{y=3, x=11, z=2}$. When $ y>4$ our last equation gives $ 7^w + 2 = 3^{y-1}s$ and $ 7^w - 2 = (3^{y-4} + 2)/s. $ with $ s$ being a positive integer. This gives us no solution since the difference of two equations is always greater than 75. So we have only one solution to this equation.