a)(2m−1)²−(2m−1)=(2m−1)(2m−1−1)(2m−1)²−(2m−1)=(2m−1)(2m−1−1)
=(2m−1)(2m−2)=2(2m−1)(m−1)=(2m−1)(2m−2)=2(2m−1)(m−1)
Ta có 2(2m-1)(m-1) chia hết cho 8 ⇔ (2m-1)(m-1) chia hết cho 4
Mà (2m-1);(m-1) = 1 ⇒ m = 4k+1 (k ∈ Z)
Vậy (2m−1)²(2m−1)(2m−1)²(2m−1) chia hết cho 8 ⇔ m = 4k+1
b)P=(xy+yz+xz)²+(x²−yz)²+(y²−yz)²+(z²−xy)²=(xy+yz+xz)²+(x²−yz)²+(y²−yz)²+(z²−xy)²
=x²y²+y²z²+x²z²+2xy²z+2x²yz+2xyz²=x²y²+y²z²+x²z²+2xy²z+2x²yz+2xyz² + x4x4 -2x²yz+y²z²2x²yz+y²z² + y4y4 -2xy²z+x²z²2xy²z+x²z² + z4z4 -2xyz²+x²z²2xyz²+x²z²
== x4x4 + y4y4 + z4z4 + 2x2y22x2y2 + 2y2z22y2z2 + 2z2x22z2x2
== (x2+y2+z2)2(x2+y2+z2)2
== 102=100102=100
