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I am trying to solve the following problem: find all solutions to the congruence $x^2 \equiv 1 \pmod {91}$.

I have solved already the congruence $x^2 \equiv 1 \pmod 7$ and $\!\!\pmod {13}$, and I am trying to use the Chinese Remainder Theorem. However, I am puzzled by how exactly to use it in this case. I do know that $x \equiv \pm 1 \pmod 7$ and $\!\!\pmod {13}$, since $7$ and $13$ are both prime.

Any help is appreciated here.

Rócherz
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letsmakemuffinstogether
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1 Answers1

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If $x\equiv 1\pmod 7$ and $x\equiv 1\pmod {13}$ then $x\equiv 1\pmod {91}$.

If $x\equiv -1\pmod 7$ and $x\equiv -1\pmod {13}$ then $x\equiv -1\pmod {91}$.

If $x\equiv 1\pmod 7$ and $x\equiv -1\pmod {13}$ then $x\equiv 64\pmod {91}$.

If $x\equiv -1\pmod 7$ and $x\equiv 1\pmod {13}$ then $x\equiv 27\pmod {91}$.

In each case, the chinese remainder theorem guarantees that the solution you found (by trial-and-error) $\pmod{91}$ is the only one.

Sonner
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    The OP wants to know how you got 1, -1, 64, and 27. – Christopher Carl Heckman Sep 24 '15 at 06:26
  • @ Sonner - Please explain why you are constructing those ``if...then..." statements as it is not clear how you got those numbers and why they are significant. – letsmakemuffinstogether Sep 24 '15 at 18:53
  • Let's consider the 3rd case for example. The procedure is this: $x\equiv 1\pmod 7$, so write $x=7a+1$. Now substitute in the second equation: $7a\equiv -2\pmod{13}$. The inverse of $7$ mod $13$ is $2$, so $a\equiv -4\equiv 9\pmod 13$, that is $a=13b+9$. Therefore $x=7(13b+9)+1=91b+64$. However for small cases like these, you can just find which one of $12,25,38,\dots$ is $\equiv 1\pmod 7$, and you'll quickly arrive to $64$. – Sonner Sep 24 '15 at 19:44