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I'd like to solve this questions without using Fermat's theorem and such. I'm stuck, since $x^2+1$ can't be factored and couldn't see how to proceed after converting $x^2 + 1 \equiv 0 \mod 11$ to $x^2 + 1 = 11k$.

Bill Dubuque
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Rodrigo
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3 Answers3

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In $\pmod{11}$ there is $11$ possibility for $x$ to be the root of $x^2+1=0$ namely $$\{0,\pm1,\pm2,\pm3,\pm4,\pm5\}$$ Substitute them in $x^2+1$ and you will see none of them satisfy the equation.

In fact for any prime $p$ with $p\equiv 3\pmod{4}$, the equation $x^2+1$ has no solution.

Qurultay
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Perhaps this is not what you might be looking for as a solution but it could be another way to solve this problem with some background in algebra.

Suppose there is a solution $x=a$ to this congruence. Then $a^2 \equiv -1 \implies a^4 \equiv 1 \pmod{11}$. This would mean the order of $a$ in the group $\mathbb{Z}_{11}^{\times}$ is $4$ but $4$ does not divide $10$ (the order of the group). Thus no such $a$ can exist.

Note this proof will work for any prime $p \equiv 3 \pmod{4}$ because $4$ does not divide the order ($p-1=4k+2$) of the multiplicative group $\mathbb{Z}_p^{\times}$.

Anurag A
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  • Is that Lagrange's Theorem? – Rodrigo May 20 '20 at 08:15
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    @Rodrigo Yes!! that is Lagrange's theorem or a simpler version that the order of an element divides the order of the group. – Anurag A May 20 '20 at 08:16
  • Thanks for pointing that out! (I'm not very familiar with the theorem, yet. But I don't think I'd spot that anyhow.) – Rodrigo May 20 '20 at 08:17
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    @Rodrigo Perhaps making you future proof :-) – Anurag A May 20 '20 at 08:19
  • I'm going to accept the other answer, as it's much more straightforward. But, I liked yours more. Thanks for being awesome, man. – Rodrigo May 20 '20 at 08:20
  • @Rodrigo If you don't yet know group theory we can whittle it away into an elementary version just as I did for Wilson's Theorem here (use $,g,h = -1,i,$ in the Remark there, where $,i^2 = -1,,$ which amounts to partitioning by the cosets of the subgroup $\langle, i,\rangle = {1, i, -1, -i},,$ as in the proof of Lagrange's theorem). See also the answer I link in the comment there. – Bill Dubuque May 21 '20 at 02:52
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    Why a down vote? Down voting without specifying the reason doesn't serve any purpose for the OP, the person answering or the community. – Anurag A May 24 '20 at 22:10
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$\mathbb Z _{11}$ is relatively small and this is a simple polynomial. You can just check if it doesn't have a root by going over all members of the field and checking if they give $0$ with the polynomial.

GSofer
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  • Well, yes that's obvious. But, I was looking for a way of doing it without testing all cases. Should have said that above. Thanks! – Rodrigo May 20 '20 at 08:15