Ok I'm stupid, any hints for this?

Here it is. First, modulo $1006$ gives $(-1)^x+(-1)^y\equiv 0\pmod{1006}$, hence, $x$ and $y$ have different parities. Next, in modulo $4$ we get for $z\geq 2$ (which is evident, as $2011>1006$), $1+3^y\equiv 0\pmod{4}$, hence, $x$ is even and $y$ is odd. Now, for $z\geq 3$, in modulo $8$, the right hand side is congruent to $0$, while the left hand side is $4$, a contradiction. Thus, only possibility is to have $z=2$. In this case, $y=1$ trivially, and it is easy to check $1006^2 - 2011= 1005^2$. The only solution is $(2,1,2)$.

grupyorum wrote: Here it is. First, modulo $1006$ gives $(-1)^x+(-1)^y\equiv 0\pmod{1006}$, hence, $x$ and $y$ have different parities. Next, in modulo $4$ we get for $z\geq 2$ (which is evident, as $2011>1006$), $1+3^y\equiv 0\pmod{4}$, hence, $x$ is even and $y$ is odd. Now, for $z\geq 3$, in modulo $8$, the right hand side is congruent to $0$, while the left hand side is $4$, a contradiction. Thus, only possibility is to have $z=2$. In this case, $y=1$ trivially, and it is easy to check $1006^2 - 2011= 1005^2$. The only solution is $(2,1,2)$. You forgot to check case z=1... If you check it you can see we have another solution as well. (x,y,z)=(1,0,1)

Apex_master wrote: If you check it you can see we have another solution as well. (x,y,z)=(1,0,1) 0 is not a positive integer.

quagmireradical wrote: Apex_master wrote: If you check it you can see we have another solution as well. (x,y,z)=(1,0,1) 0 is not a positive integer. Positive integers mean natural numbers +0 Dont trust internet on everything

Apex_master wrote: quagmireradical wrote: Apex_master wrote: If you check it you can see we have another solution as well. (x,y,z)=(1,0,1) 0 is not a positive integer. Positive integers mean natural numbers +0 Dont trust internet on everything Em... where do you get that information from ?! It is obvious that positive excludes 0, non-negative includes 0. Dont trust internet on everything

parmenides51 wrote: Solve in positive integers the equation $1005^x + 2011^y = 1006^z$. As $1006|1005^x+(1005+1006)^y$ and as $(2011)^y=1005^y+1006*m\implies 1006|1005^x+1005^y$ hence $(x, y)$ are not both odd or even. Now if $z\geq 3$ we have $1006^z\equiv 0\mod 8$ and $1005^x+ 2011^y\equiv (5)^x+(3)^y\not\equiv 0\mod 8$ so $z<3$ and also $z\geq x, z>y$ Case 1-: $z=2$ so $2\geq x, 2>y$ clearly $x=2, y=1$ only works in this case. Case 2-: $z=1$ so $1\geq x, 1>y$ clearly $x=1, y=0$ Is only Solution in this case. Case 3-: $z=0$ so $0\geq x, 0>y$ clearly as $y$ becomes negative so this case has no solution. Hence the only solutions are $(x,y,z)=(1,0,1), (2,1, 2)\blacksquare$

mod 1006 and mod 8