(a^2+b^2-c^2)^2
=a^4+b^4+c^4+2a^2b^2-2b^2c^2-2c^2a^2
=2a^2b^2,
a^2+b^2-c^2=±√2ab,
cosC=(a^2+b^2-c^2)/2ab=±√2/2,
C=45°或135°
a^4+b^4+c^4=2c^2(a^2+b^2)
===> (a^2+b^2)^2-2a^2b^2+c^4-2c^2(a^2+b^2)=0
由余弦定理得到：a^2+b^2-c^2=2abcosC
所以：a^2+b^2=c^2+2abcosC
===> (c^2+2abcosC)^2-2a^2b^2+c^4-2c^2(c^2+2abcosC)=0
===> c^4+4abc^2cosC+4a^2b^2cos^2C-2a^2b^2+c^4-2c^4-4abc^2cosC=0
===> 4a^2b^2cos^2C=2a^2b^2
===> cos^2C=1/2
===> cosC=±√2/2
===> C=45°或者135°