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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!

Hawk
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    Note that $14$ is a multiple of $7$. What does that tell you about a putative integer zero $k$ of the polynomial? – Daniel Fischer Jan 12 '14 at 16:56
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    Sounds like Daniel Fischer is riffing a theme by Eisenstein? – Jyrki Lahtonen Jan 12 '14 at 17:00
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    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. – Hawk Jan 12 '14 at 17:01

2 Answers2

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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.

Community
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Bill Dubuque
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    Is there a result that says $7| (n^2+1)$ is not true? – Tapi Nov 22 '20 at 18:43
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    @Tapi $7\mid n^2+1\iff \bmod 7!:\ n^2\equiv -1,,$ so we need only check that $-1$ is not a square $\bmod 7$, i.e. $,x^2+1,$ has no roots $\bmod 7,,$ verifiable either by brute force checking the few cases, or, more generally, by Euler's criterion or quadratic reciprocity - see here. – Bill Dubuque Nov 23 '20 at 01:56
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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.

Xucheng Zhang
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