a^4+b^4+C^4+d^4＝4abcd
a^4+b^4+c^4+d^4-4abcd=0
( a^4+b^4-2a^2b^2)+(c^4+d^4-2c^2d^2)+(-4abcd+2a^2b^2+2c^2d^2)=0
(a^2-b^2)^2+(c^2-d^2)^2+2(ab-cd)^2=0
又a、b、c、d为正,
即有a=b=c=d.