# Three variables

### Three variables

Prove that the equation
$x^{2}+y^{2}+z^{2}=2^{m}$
has no natural solutions $x,y,z$ for any natural number m. (the problem for m=9 was proposed on second round of the bulgarian mathematical olympiad for seven grade in 2005)

MM

Posts: 78
Joined: Tue Jul 22, 2008 7:36 am
Location: Bulgaria

### Re: Three variables

MM wrote:Prove that the equation
$x^{2}+y^{2}+z^{2}=2^{m}$
has no natural solutions $x,y,z$ for any natural number m. (the problem for m=9 was proposed on second round of the bulgarian mathematical olympiad for seven grade in 2005)

m=1 it's crear no solution, because lhs > 2
m>1 => 4/rhs => 4/lhs. but lhs mod 4 is 0,1,2 or 3 and it's 0 only if every summand is 0 mod 4 => 2/x, 2/y, 2/z. x=2a, y=2b, z=2c => a^2+b^2+c^2=2^(m-2). so the same problem if m>2 so continuing at last we get the problem a^2+b^2+c^2 = m-2k = 1 or 2. if m is even than no solution. else again no solution
martin123456

Posts: 12
Joined: Tue Nov 17, 2009 10:15 am