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Following the suggestion by J.G., for the induction step we assume true that

  • $Q(n) = x^{n} - y^{n} = p_{n-1}(x)(x-y)$$Q(n) = x^{n} - y^{n} = p_{n-1}(x)\cdot (x-y)$

were $p_{n-1}(x)$ is a polynomial of degree $n-1$ then

  • $Q(n+1) = x^{n+1} - y^{n+1} = x(x^n-y^n)+y^n(x-y)=x\,\color{red}{p_{n-1}(x)(x-y)}+y^n(x-y)=p_n(x)(x-y)$$Q(n+1) = x^{n+1} - y^{n+1} = x\cdot(x^n-y^n)+y^n(x-y)=x\cdot\,\color{red}{p_{n-1}(x)\cdot(x-y)}+y^n(x-y)=p_n(x)\cdot(x-y)$

Following the suggestion by J.G., for the induction step we assume true that

  • $Q(n) = x^{n} - y^{n} = p_{n-1}(x)(x-y)$

then

  • $Q(n+1) = x^{n+1} - y^{n+1} = x(x^n-y^n)+y^n(x-y)=x\,\color{red}{p_{n-1}(x)(x-y)}+y^n(x-y)=p_n(x)(x-y)$

Following the suggestion by J.G., for the induction step we assume true that

  • $Q(n) = x^{n} - y^{n} = p_{n-1}(x)\cdot (x-y)$

were $p_{n-1}(x)$ is a polynomial of degree $n-1$ then

  • $Q(n+1) = x^{n+1} - y^{n+1} = x\cdot(x^n-y^n)+y^n(x-y)=x\cdot\,\color{red}{p_{n-1}(x)\cdot(x-y)}+y^n(x-y)=p_n(x)\cdot(x-y)$
Source Link
user
  • 156.6k
  • 12
  • 80
  • 147

Following the suggestion by J.G., for the induction step we assume true that

  • $Q(n) = x^{n} - y^{n} = p_{n-1}(x)(x-y)$

then

  • $Q(n+1) = x^{n+1} - y^{n+1} = x(x^n-y^n)+y^n(x-y)=x\,\color{red}{p_{n-1}(x)(x-y)}+y^n(x-y)=p_n(x)(x-y)$