<s>[INST] Prove that there do not exist positive integers $a,b$ such that $p \mid a+b+1,a^2+b^2+1,a^3+b^3+1$ where $p > 3$ is a prime 
Prove that if $p > 3$ is a prime then there do not exist positive integers $a,b$ such that \begin{align*}a+b+1 &\equiv 0 \pmod{p}\\a^2+b^2+1 &\equiv 0 \pmod{p}\\a^3+b^3+1 &\equiv 0 \pmod{p}.\end{align*}

We have $a+b \equiv a^2+b^2 \equiv a^3+b^3 \pmod{p}$. Thus $a(a-1)+b(b-1) \equiv 0 \pmod{p}$ and $a^2(a-1)+b^2(b-1) \equiv 0 \pmod{p}$.Then $$a(a-1)+b(b-1) \equiv a^2(a-1)+b^2(b-1) \pmod{p}.$$ How can we continue from here to show that the equation doesn't have solutions in positive integers?
 [/INST] Let $p\gt 3$ be a prime. In the field $\mathbb F_p$ we have $$\begin{align*}a+b=-1\\a^2+b^2=-1\\a^3+b^3=-1\end{align*}$$ It follows from the first equation $$a^2+b^2+2ab=1 \Rightarrow ab=1$$
$$a^3+b^3+3ab(a+b)=-1 \Rightarrow 3ab(a+b)=0 \Rightarrow ab=0$$ Thus $$ 0=-1\text{ contradiction}.$$
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