即(2p+1)|(2P-1)例如:23|(2-1);11=4×2+347|(2-1);23=4×5+3167|(2-1);83=4×20+32、p=2n×32+1,,则(6p+1)|(2P-1),例如:223|(2-1);37=2×2×3×3+1439|(2-1);73=2×2×2×3×3+13463|(2-1);577=2×2×2×2×2×2×3×3+13、p=2n×3m×5s-1,则(8p+1)|(2p-1)例如;233|(2-1);29=2×3×5-11433|(2-1);179=2×2×3×3×5-11913|(2-1);239=2×2×2×2×3×5-1例1:分解因式(1+y)2-2x2(1+y2)+x(1-y)2.解:原式=(1+y)2+2(1+y)x2(1-y)+x(1-y)2-2(1+y)x2(1-y)-2x2(1+y2)(补项)=[(1+y)+x2(1-y)]2-2(1+y)x2(1-y)-2x2(1+y2)(完全平方)=[(1+y)+x2(1-y)]2-(2x)2=[(1+y)+x2(1-y)+2x][(1+y)+x2(1-y)-2x]=(x2-x2y+2x+y+1)(x^2-x2y-2x+y+1)=[(x+1)2-y(x2-1)][(x-1)2-y(x2-1)]=[(x+1)2-y(x+1)(x-1)][(x-1)2-y(x+1)(x-1)]=(x+1)(x+1-xy+y)(x-1)(x-1-xy-y).例2:求证:对于任何整数x,y,下式的值都不会为33:x+3xy-5x3y2-15x2y3+4xy+12y.解:原式=(x+3xy)-(5x3y2+15x2y3)+(4xy+12y)=x(x+3y)-5x2y2(x+3y)+4y(x+3y)=(x+3y)(x-5x2y2+4y)=(x+3y)(x2-4y2)(x2-y2)=(x+3y)(x+y)(x-y)(x+2y)(x-2y).当y=0时,原式=x不等于33;当y不等于0时,x+3y,x+y,x-y,x+2y,x-2y互不相同,而33不能分成四个以上不同因数的积,所以原命题成立。例3:△ABC的三边a、b、c有如下关系式:-c2+a2+2ab-2bc=0,求证:这个三角形是等腰三角形。分析:此题实质上是对关系式的等号左边的多项式进行因式分解。证明:∵-c2+a2+2ab-2bc=0,∴(a+c