\[x^2+y^2+w^2+z^2=2xywz\]
have solutions?The right hand side is even, so on the left hand side 0,2 or 4 of
\[x, \; y, \; w, \; z\]
must be odd. If they are all odd, then the square of any odd number equals 1 modulus 4 so the left hand side equals 0 modulus 4 and the right hand side equals 2 modulus 4; If two are odd then the left hand side equals 2 modulus 4, but the right hand side equals 0 modulus 4. Hence all must be even.Let
\[x=2x', \; y=2y', \; w=2w', z=2z'\]
then the equation becomes\[4x'^2+4y'^2+4w'^2+4z'^2=2(2x')(2y')(2w;)(2z')=32x'y'w'z'\]
\[x'^2+y'^2+w'^2+z'^2=8x'y'w'z'\]
.If the Diophantine equation
\[x'^2+y'^2+w'^2+z'^=8kx'y'w'z'\]
has solutions, then wince the right hand side is even, 0, 2 or 4 of \[x', \; y', \; w', \; z'\]
on the left hand side must be even. The square of any odd integer is congruent modulo 8 to 1 so if all of \[x', \; y', \; w', \; z'\]
are odd then the left hand side is congruent modulo 8 to 4, and if only two are the left hand side is congruent modulo 8 to 2 or 6, but the right hand side is congruent modulo 8 to 0, so all are even.Let
\[x'=2x'', \; y'=2y'', \; w'=2w', z'=2z''\]
then the equation becomes\[4x''^2+4y''^2+4w''^2+4z''^2=8k(2x'')(2y'')(2w'')(2z'')=128kx''y''w''z''\]
\[x''^2+y''^2+w''^2+z''^2=32kx''y''w''z''=8(4k)x''y''w''z''\]
.We cannot continue to reduce this equation in the same way, so no solution exists.