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Determine all pairs of positive integers $(x,n)$ which satisfy the condition $$x^3+2x+1=2^n.$$

My work so far:

No solution exists for $n=1$. For $n=2$ we get $x=1$.

We show that no solutions exist for $n\ge3$. Suppose that $n \ge 3$. Obviously, $x$ is odd. Then $x^2+2\equiv 3 \pmod 8$. As with the original equation $x(x^2+2)\equiv -1 \pmod 8$, then $x\equiv 5\pmod 8$.

I can not get a contradiction.

Apass.Jack
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Roman83
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  • No solution exists for $n=1$. For $n=2$ we get $x=1$. –  Feb 19 '16 at 13:54
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    If you've solved the problem yourself, consider posting an answer, both for future reference for those with similar problems, as well as preventing it from being stuck in the "unanswered" tab. –  Feb 19 '16 at 14:06

1 Answers1

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Suppose $n \ge 3$.

Note that $x \equiv 5 \pmod 8$.

Since $x^3+2x+1 \equiv 3x+1 \equiv 1 \pmod 3$ by Fermat's little theorem, $2^{n}$ divided by $3$ has a remainder of $1$. This implies that $n$ is a even number. Let $n=2k$.

By adding $2$ to each side, the equation can be factorized so that $$(x+1)(x^2-x+3)=2^{2k}+2.$$

Since $x^2-x+3 \equiv 23 \equiv 7 \pmod 8$, there exists a prime number $p$ where $p \equiv 5, 7\pmod8$ such that $p$ divides $x^2-x+3$. (If there is no such $p$, then $x^2-x+3 \equiv 3^{k} \pmod 8$. But $3^{k} \not \equiv 7 \pmod 8$). Since $x^2-x+3$ is a factor of $2^{2k}+2$, $p$ divides $2^{2k}+2$.

That means $(2^{k})^2 \equiv -2 \pmod p$. But $-2$ is a non-quadratic residue of $p$ since $p \not\equiv 1,3\pmod 8$. A contradiction.

Apass.Jack
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S.C.B.
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