思路都是比較兩邊的根.
1.x = 1是(x-1)·f(x+1)的根,所以也是(x+2)·f(x)的根.
但其不是x+2的根,故其為f(x)的根,f(1) = 0.
同理,由x = -2是(x+2)·f(x)的根,可得f(-1) = 0.
繼而將x = 0代入左端,得f(0) = 0.
由f(x)有根0,1,-1,可設(shè)f(x) = x(x-1)(x+1)·g(x).
代入等式得x(x-1)(x+1)(x+2)·g(x+1) = x(x-1)(x+1)(x+2)·g(x).
既有g(shù)(x) = g(x+1),而g(x)是多項(xiàng)式,只有g(shù)(x) = c.
于是f(x) = cx(x-1)(x+1),其中c為非零實(shí)數(shù).
2.首先f(wàn)(x) = 0與f(x) = 1顯然滿(mǎn)足要求,以下只討論次數(shù)大于1的多項(xiàng)式.
由f(x²) = f(x)f(x+1),若a是f(x)的根,則a也是f(x²)的根,也即a²是f(x)的根.
于是a,a²,(a²)²,((a²)²)²,...都是f(x)的根.
但若f(x)非零,只有有限個(gè)根,存在m < n使a^m = a^n,于是a^m·(a^(n-m)-1) = 0.
有a = 0,或a是單位根(某個(gè)正整數(shù)次冪為1).
f(x)的根只能為0或單位根,于是f(x²)的根也只能為0或單位根.
由f(x²) = f(x)f(x+1),f(x+1)的根也只能為0或單位根.
而a是f(x)的根當(dāng)且僅當(dāng)a-1是與f(x+1)的根.
若a與a-1都是單位根,設(shè)b是a的復(fù)共軛,有ab = |a|² = 1,(a-1)(b-1) = |a-1|² = 1.
可解得a = (1±√3i)/2,記α = (1+√3i)/2,β = (1-√3i)/2.
若a-1 = 0,則a = 1.
于是f(x)的根只能為0,1,α,β,f(x+1)的根只能為-1,0,α-1,β-1.
若α是f(x)的根,則√α是f(x²)的根,但√α不是f(x)或f(x+1)的根,與f(x²) = f(x)f(x+1)矛盾.
故α不是f(x)的根,同理β也不是f(x)的根,故f(x)的根只能為0,1.
設(shè)f(x) = c·x^m·(x-1)^n.
有f(x²) = c·(x²)^m·(x²-1)^n = c·x^(2m)·(x-1)^n·(x+1)^n,f(x+1) = c·(x+1)^m·x^n.
代入等式得c·x^(2m)·(x-1)^n·(x+1)^n = c²·x^(m+n)·(x-1)^n·(x+1)^m.
當(dāng)c ≠ 0,等式成立當(dāng)且僅當(dāng)m = n,c = 1.
故f(x) = x^m·(x-1)^m.
滿(mǎn)足條件的多項(xiàng)式只有0,1和x^m·(x-1)^m.