Find all positive integers $x,y,z$ such that $7^x + 13^y = 8^z$
Problem
Source: 2013 Romania JBMO TST 3.2
Tags: Diophantine equation, diophantine, number theory
jelena_ivanchic
02.03.2021 10:16
This forum has really nice and cute problems! Thank you so much parmenidies51, for collecting so nice problems!
Note that $z>0$. So taking $\mod 8,$ we get $x$ is odd and $y$ is even. If $y=0\implies $ by zsig, the only solution is $(x,z)=(1,1).$
If $y>0 \implies x \equiv 3 \mod 12, z \equiv 3 \mod 4.$ Letting $x=3k \implies 2^{3z}-7^{3k}=13^y \implies (2^z-7^k)(7^{2k}+2^z\cdot 7^k+2^{2z})=13^y.$
Letting $d=(2^z-7^k,7^{2k}+2^z\cdot 7^k+2^{2z})\implies d|2\cdot 2^z\cdot 7^k\implies 2^z-7^k =1.$ By zsig, the only sol is $(z,k)=(3,1) \implies y=2.$
So the solutions are $\boxed{(x,y,z)=(1,0,1),(3,2,3)}$
P2nisic
07.03.2021 00:26
parmenides51 wrote: Find all positive integers $x,y,z$ such that $7^x + 13^y = 8^z$ This is very strange for Roumania we can solve the following: Find all nonegative integers $x,y,z$ such that$7^x+13^y=2^z$ If $x,y>0$ we have Taking $mod 7$ we have $z=3a$ Taking $mod13$ we have $x=3b$ Now rewrite the equation as:$13^y= (2^a-7^b)(7^{2b}+2^a\cdot 7^b+2^{2a})$ It is well know that$(2^a-7^b,7^{2b}+2^a\cdot 7^b+2^{2a})=1or3$ but $3$ doesn't decid13. So we have that $2^a=7^b+1$ taking $mod16$ we have that $(a,b):(1,0),(3,1)$ Solution $(x,y,z):(3,2,9)$ If $y=0$ we have: $7^x+1=2^z$ Solution $(x,y,z):(0,0,1)(1,0,3)$ If $x=0$ we have $1+13^y=2^z$ If $y>0$ we have contradiction by $mid3,13$ So all the solution are:$\boxed{(x,y,z):(0,0,1),(1,0,3),(3,2,9)}$
jelena_ivanchic
08.03.2021 10:52
P2nisic wrote: parmenides51 wrote: Find all positive integers $x,y,z$ such that $7^x + 13^y = 8^z$ This is very strange for Roumania we can solve the following: Find all nonegative integers $x,y,z$ such that$7^x+13^y=2^z$ If $x,y>0$ we have Taking $mod 7$ we have $z=3a$ Taking $mod13$ we have $x=3b$ Now rewrite the equation as:$13^y= (2^a-7^b)(7^{2b}+2^a\cdot 7^b+2^{2a})$ It is well know that$(2^a-7^b,7^{2b}+2^a\cdot 7^b+2^{2a})=1or3$ but $3$ doesn't decid13. So we have that $2^a=7^b+1$ taking $mod16$ we have that $(a,b):(1,0),(3,1)$ Solution $(x,y,z):(3,2,9)$ If $y=0$ we have: $7^x+1=2^z$ Solution $(x,y,z):(0,0,1)(1,0,3)$ If $x=0$ we have $1+13^y=2^z$ If $y>0$ we have contradiction by $mid3,13$ So all the solution are:$\boxed{(x,y,z):(0,0,1),(1,0,3),(3,2,9)}$ Are you sure about your solutions? Clearly $7^0+13^0\ne 8^1$ ? Or did I read something wrong? I think here's the mistake, here:- Quote: $7^x+1=2^z$ It should be $7^x+1=2^{3z}.$ The solution you got are solutions of $7^x+1=2^{3z}.$ And you got $(x,y,3z)=(0,0,1),(1,0,3),(3,2,9)$ as solutions, clearly $1$ is not divisible by $3$ hence first solution is wrong. So you get $(x,y,3z)=(1,0,3),(3,2,9)\implies (x,y,z)=(1,0,1),(3,2,3),$ which is the same solution set I got! @below oh! my bad!
P2nisic
08.03.2021 13:54
jelena_ivanchic wrote: P2nisic wrote: parmenides51 wrote: Find all positive integers $x,y,z$ such that $7^x + 13^y = 8^z$ This is very strange for Roumania we can solve the following: Find all nonegative integers $x,y,z$ such that$7^x+13^y=2^z$ If $x,y>0$ we have Taking $mod 7$ we have $z=3a$ Taking $mod13$ we have $x=3b$ Now rewrite the equation as:$13^y= (2^a-7^b)(7^{2b}+2^a\cdot 7^b+2^{2a})$ It is well know that$(2^a-7^b,7^{2b}+2^a\cdot 7^b+2^{2a})=1or3$ but $3$ doesn't decid13. So we have that $2^a=7^b+1$ taking $mod16$ we have that $(a,b):(1,0),(3,1)$ Solution $(x,y,z):(3,2,9)$ If $y=0$ we have: $7^x+1=2^z$ Solution $(x,y,z):(0,0,1)(1,0,3)$ If $x=0$ we have $1+13^y=2^z$ If $y>0$ we have contradiction by $mid3,13$ So all the solution are:$\boxed{(x,y,z):(0,0,1),(1,0,3),(3,2,9)}$ Are you sure about your solutions? Clearly $7^0+13^0\ne 8^1$ ? Or did I read something wrong? I think here's the mistake, here:- Quote: $7^x+1=2^z$ It should be $7^x+1=2^{3z}.$ The solution you got are solutions of $7^x+1=2^{3z}.$ And you got $(x,y,3z)=(0,0,1),(1,0,3),(3,2,9)$ as solutions, clearly $1$ is not divisible by $3$ hence first solution is wrong. So you get $(x,y,3z)=(1,0,3),(3,2,9)\implies (x,y,z)=(1,0,1),(3,2,3),$ which is the same solution set I got! I had solve this $7^x+13^y=2^z$ and not this $7^x+13^y=8^z$