Three variables

by MM » Sun Dec 28, 2008 1:33 pm

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)

by **martin123456** » Wed Nov 25, 2009 5:14 am

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

Three variables

Three variables

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