■19四乗数定理(その38)
6(a^2+b^2+c^2+d^2)^2
=(a+b)^4+(a−b)^4+(c+d)^4+(c−d)^4
+(a+c)^4+(a−c)^4+(b+d)^4+(b−d)^4
+(a+d)^4+(a−d)^4+(b+c)^4+(b−c)^4
について,右辺を展開してみると
2(a^4+6a^2b^2+b^4)+2(c^4+6c^2d^2+d^4)
+2(a^4+6a^2c^2+c^4)+2(b^4+6b^2d^2+d^4)
+2(a^4+6a^2d^2+d^4)+2(b^4+6b^2c^2+c^4)
=6(a^4+b^4+c^4+d^4)+12(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)=左辺
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6(x^2+y^2+z^2+w^2)
=(x−y)^2+(z−w)^2+(x+y)^2+(z+w)^2
+(x−z)^2+(y−w)^2+(x+z)^2+(y+w)^2
+(x−w)^2+(y−z)^2+(x+w)^2+(y+z)^2
について,右辺を展開してみると
2(x^2+y^2)+2(z^2+w^2)
+2(x^2+z^2)+2(y^2+w^2)
+2(x^2+w^2)+2(y^2+z^2)
=6(x^2+y^2+z^2+w^2)
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