Let $p \geq 3$ be an integer and $\alpha, \beta$ be the roots of $x^{2}-(p+1) x+1=0$ using mathemati

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Let $p \geq 3$ be an integer and $\alpha, \beta$ be the roots of $x^{2}-(p+1) x+1=0$ using mathematical induction

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Assume that $\alpha^{k}+\beta^{k}$ be an integer not divisible by $p$ where $k$ is any integer $>2 \ldots$ (A)\ Now $\alpha+\beta=p+1$ and $\alpha \beta=1$ as $\alpha, \beta$ be the roots of\ $x^{2}-(p+1) x+1=0$.\ We have $(\alpha+\beta)\left(\alpha^{k}+\beta^{k}ight)$\ $=\alpha^{k+1}+\beta^{k+1}+\alpha \beta\left(\alpha^{k-1}+\beta^{k-1}ight)$\ $	herefore \quad \alpha^{k+1}+\beta^{k+1}$\ $=(\alpha+\beta)\left(\alpha^{k}+\beta^{k}ight)-\alpha \beta\left(\alpha^{k-1}+\beta^{k-1}ight)$\ or $\quad \alpha^{k+1}+\beta^{k+1}$\ $=(p+1)\left(\alpha^{k}+\beta^{k}ight)-\left(\alpha^{k-1}+\beta^{k+1}ight)$\ As the right hand of this equation is an integer by virtue of (A) as also ( $p+1$ )\ $	herefore \quad \alpha^{k+1}+\beta^{k+1}$ is also an integer.\ Now, $\alpha+\beta=p+1$ so $\alpha+\beta$ is not divisible\ by $p$\ Also $(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2 \alpha \beta$ $$\begin{equation*}=p^{2}+2 p+1 	ag{I} \end{equation*}$$ or $\alpha^{2}+\beta^{2}=p^{2}+2 p+1-2 \alpha \beta$\ $=p^{2}+2 p-1$\ $	herefore \quad \alpha^{2}+\beta^{2}$ is also not divisible by $p$\ Again $\alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)$ $$\begin{equation*}=(p+1)^{3}-3(p+1) 	ag{II} \end{equation*}$$ $=p^{3}+3 p^{2}-2$ As $p \geq 3, p^{3}+3 p^{2}-2$ is also not divisible by $p$\ Let $f(k)=\alpha^{k}+\beta^{k}$ $$\begin{equation*}	herefore \quad f(k+1)=\alpha^{k+1}+\beta^{k+1} 	ag{III} \end{equation*}$$ $=(\alpha+\beta)\left(\alpha^{k}+\beta^{k}ight)-\alpha \beta\left(\alpha^{k-1}+\beta^{k-1}ight)$\ or $\quad f(k+1)=(p+1) f(k)-f(k-1)$\ $	herefore \quad f(k+2)=(p+1) f(k+1)-f(k)$\ or $f(k+2)=(p+1)[(p+1) f(k)$\ $-f(k-1)]-f(k)$\ $=f(k) 	imes\left(p^{2}+2 pight)-(p+1) f(k-1)$\ $=\left[\left(p^{2}+2 pight) f(k)-p f(k-1)ight]-f(k-1)$\ $	herefore \quad$ If $f(k-1)$ is not a multiple of $p$, so also is $f(k+2)$. Using (I), (II) and (III) we see that\ $f(4), f(5) \ldots$. are not multiple of $p$.\ $	herefore \quad f(n)=\alpha^{n}+\beta^{n}$ is not a multiple of $p$.\