Solve in positive integers the equation 1005x+2011y=1006z.
Problem
Source: JBMO 2011 Shortlist N1
Tags: JBMO, number theory, algebra, integer equation, Diophantine equation, modular arithmetic
23.12.2018 00:51
Ok I'm stupid, any hints for this?
23.12.2018 01:35
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).
25.09.2020 14:32
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)
25.09.2020 15:09
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.
13.10.2020 11:13
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
13.10.2020 11:20
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
13.10.2020 19:51
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
03.02.2025 05:58
mod 1006 and mod 8