<s>[INST] Show that the equation $x^3+7x-14(n^2+1)$ has no integral root for any integer $n$. Show that the equation
$$x^3+7x-14(n^2+1)=0$$
has no integral root for any integer $n$.
My work:
I consider the contraposition that there are integer roots.
Assume that the roots are $\alpha,\beta,\gamma$
We have, $\alpha\beta\gamma=14(n^2+1)$
We also have, $\alpha\beta+\alpha\gamma+\beta\gamma=7$
Since $n^2\ge 1$, we get, $\alpha\beta\gamma\ge 28$
Now, I have no idea! Please help!
 [/INST] Consider $x^3+7x-14(n^2+1)\equiv0 \mod (7)$, we get $x^3\equiv0 \mod(7)$ and therefore $x\equiv0 \mod (7)$.
Assume $x=7t$, then $(7t)^3+7\cdot 7t-14(n^2+1)=0$. i.e., $7^2t^3+7t-2(n^2+1)=0$. Similar, $7^2t^3+7t-2(n^2+1)\equiv0\mod (7)$, one has $2(n^2+1)\equiv0 \mod (7)$, which is impossible for integer $n$. So the equation has no integral roots.
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