Show that the equation
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!

  • 1
    $\begingroup$ Note that $14$ is a multiple of $7$. What does that tell you about a putative integer zero $k$ of the polynomial? $\endgroup$ – Daniel Fischer Jan 12 '14 at 16:56
  • $\begingroup$ Sounds like Daniel Fischer is riffing a theme by Eisenstein? $\endgroup$ – Jyrki Lahtonen Jan 12 '14 at 17:00
  • $\begingroup$ So, you mean, $\alpha\beta\gamma=2(\alpha\beta+\alpha\gamma+\beta\gamma)(n^2+1)$, which says that 1 of the 3 roots are even. But, I cannot conclude something more. $\endgroup$ – Hawk Jan 12 '14 at 17:01

Hint $\,\ 7\mid x^3\Rightarrow\ 7\mid x.\ $ Substitute $\, x = 7y,\,$ cancel $7,\,$ then deduce $\,7\mid n^2+1,\,$ contradiction.

Remark $\ $ Generally, $ $ prime $\,p\mid x^k\!+pnx\, \Rightarrow\, p\mid x^k \Rightarrow p\mid x \Rightarrow p^2\mid x^k\!+pnx\ $ if $\ k\ge 2$

This may be viewed as a special case of Eisenstein's criterion.


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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