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The question is from a multiple choice problem:

How many elements $x$ in the field ${\Bbb Z}_{11}$ satisfy the equation $x^{12}-x^{10}=2$?
A.1 B.2 C.3 D.4 E.5

I know that one may need Fermat's little theorem to solve $x^{10}(x^2-1)\equiv 2\pmod{11}$.

How should I go on?

  • I frankly don't understand your question. How do you show that the equation $x^{12}-x^{10}=2$ is equivalent to $x^{10}(x^2-1)=2$, when you are looking for solutions in the real numbers? The same principles apply. – Jyrki Lahtonen Jun 16 '11 at 06:10
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    Or is the problem that you don't know that working in ${\bf Z}_{11}$ is the same thing as working mod 11? – Gerry Myerson Jun 16 '11 at 07:07
  • @Gerry: Yes, you are right. I don't understand why working in ${\bf Z}_{11}$ is the same thing as working mod 11. –  Jun 16 '11 at 14:46
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    @Jack: If you post a separate question, I (or others) can give you the answer of why "working in $\mathbb{Z}_m$" is essentially the same as "working with congruences modulo $m$". But the comment boxes are too narrow to contain it. – Arturo Magidin Jun 16 '11 at 16:18
  • @Arturo: Thanks for your patience. I have added the question here. –  Jun 16 '11 at 16:38
  • @Jack: I apologize for coming thru a bit harsh with the first comment. In retrospect I should have recognized what your problem was. Glad that it has now been identified and remedied. This will quickly become a deep-rooted second nature. So much of second nature that you no longer see the difference! – Jyrki Lahtonen Jun 16 '11 at 19:43
  • @Arturo "But the comment boxes are too narrow to contain it." I take that as alluding to the other Fermat's theorem :-) – Srivatsan Aug 02 '11 at 06:34
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    @Srivatsan: Only unconsciously. – Arturo Magidin Aug 02 '11 at 08:12

1 Answers1

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Fermat's Little Theorem tells you that $x^{11}\equiv x\pmod{11}$, and that if $x$ is not congruent to $0$ modulo $11$, then $x^{10}\equiv 1\pmod{11}$. Since $x=0$ is certainly not a solution, we may restrict our attention to nonzero $x$; so $x^{12} = x^{10}x^2 \equiv x^2\pmod{11}$, and $x^{10}\equiv 1 \pmod{11}$.

Then this becomes $x^2-1\equiv 2\pmod{11}$, or $x^2\equiv 3\pmod{11}$. There are either two or no solutions, and since $0$ is not an option, the answer is B.

The fact that $x^{12}-x^{10}\equiv x^{10}(x^2-1)\pmod{11}$ is simply because $11$ divides $(x^{12}-x^{10}) - x^{10}(x^2-1)$, no matter what (integer) $x$ is.

Added. I don't understand your added question. $a$ is a solution to $x^{10}(x^2-1)\equiv 2\pmod{11}$ if and only if it is a solution to $x^{12}-x^{10}\equiv 2 \pmod{11}$, simply because $$x^{12}-x^{10} = x^{10}(x^2-1)$$ for any integer (for any complex number, even!). Things that are equal remain equal modulo any equivalence relation (such as congruences), because equivalence relations are always reflexive.

Apropos of Nothing. You can determine if there are solutions or not (if you didn't have the answers in front of you) either by simply testing the numbers modulo $11$, or by using quadratic reciprocity. Both $3$ and $11$ are congruent to $3$ modulo $4$; since $11\equiv 2\pmod{3}$ is not a square modulo $3$, then $3$ is a square modulo $11$.

Arturo Magidin
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  • It's my ignorance that I didn't understand why working in ${\bf Z}{11}$ is the same thing as working mod 11. I understood the definition of congruence superficially. "$a=b$ in the field ${\bf Z}{m}$" is the same thing as "$a\equiv b\pmod{m}$". Right? Now I see. –  Jun 16 '11 at 15:49
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    @Jack: Except that $\mathbb{Z}_m$ is a field if and only if $m$ is a prime. The elements of the ring $\mathbb{Z}_m$ are equivalence classes of integers, where the equivalence relation is "congruent modulo $m$, so indeed, saying "$a=b$ in the ring $\mathbb{Z}_m$" is in fact the exact same thing as saying $a\equiv b \pmod{m}$, if you use an integer $a$ and $b$ to represent the congruence class modulo $m$. – Arturo Magidin Jun 16 '11 at 16:17