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I have the Answer to this Question :
How many solutions does $x^2 \equiv 1 \mod 680$ have?
By the Chinese Remainder theorem, we get $16$ solutions to this.

We do as follows:
$x^2 \equiv 1 \mod 8$ has solutions namely $1,3,5,7$ ;
$x^2 \equiv 1 \mod 5$ has solutions namely $1,-1$ ; and
$x^2 \equiv 1 \mod 17$ has solutions $1,-1$.
Hence the number of solutions is $4 \times 2 \times 2 = 16$.

DOUBT :

In case of $\mod 5$ , we counted $\pm 1$
In case of $\mod 17$ , we counted $\pm 1$
Why did we not count the negative integers in case of $\mod 8$ ie $-1,-3,-5,-7$ ?

Why did we ignore those negative Solutions for $8$ , while counting it for $5$ & $17$ ?

Prem
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FriedSpies
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    $-1 = 7 \mod 8$ and so on. – Amateur_Algebraist Dec 10 '23 at 08:04
  • but we have $x^2$ so will it become $(-1)^1=1=1mod(8)$ – FriedSpies Dec 10 '23 at 08:10
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    This is a duplicate of this oldie and several others. I won't vote to close it right away, because my vote would be immediately binding, and I'm not sure that's the best target. Even though it is listed in the list of FAQs. – Jyrki Lahtonen Dec 10 '23 at 08:33
  • Handling the prime factor $2$ of the modulus is the trickiest of the lot. – Jyrki Lahtonen Dec 10 '23 at 08:35
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    Anyway, for the purposes of counting solutions, two solutions are considered to be the same, if they are congruent w.r.t. $680$. CRT says that two solutions are congruent modulo $680$ if and only if they are congruent modulo $16$, $5$ and $17$. That's why you can ignore the rest of the solutions modulo $8$. After all $-1\equiv 7\pmod8$ etc. – Jyrki Lahtonen Dec 10 '23 at 08:37
  • Should read $8,5,17$ in the preceding comment. $16$ is not a factor of $680$. – Jyrki Lahtonen Dec 10 '23 at 09:32
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    Congruent roots are considered the same in modular arithmetic, i.e. we count the roots in any complete residue system, see here in the linked dupe. So here $!\bmod 8!:\ {-}1\equiv 7,\ {-}3\equiv 5,\ldots\ \ $ – Bill Dubuque Dec 10 '23 at 10:55
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    e.g. $,x+1\equiv 0\pmod{!5},$ has exactly one root in any complete residue system. The root is $,4,$ in the least natural residue system $,{0,1,2,3,4},,$ but $,{-1},$ in the least magnitude (balanced) system $,{0,\pm1,\pm2}.\ \ $ – Bill Dubuque Dec 10 '23 at 11:17
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    Dups are available & Solutions can be attempted though known ways , though that is not the Core Concern here. OP knows the Answer & a Solution. OP wants clarification for the DOUBT highlighted. I am not very sure whether that Dup is available. – Prem Dec 10 '23 at 11:42
  • BTW , the DOUBT clarification is very simple : It is immaterial whether we choose Negative Solutions or not : The count will remain unchanged. Solution list is $1, 69, 101, 169, 171, 239, 271, 339, 341, 409, 441, 509, 511, 579, 611, 679$ – Prem Dec 10 '23 at 15:21
  • My contention is only that the Dup "Question" is not about that. No worries , @BillDubuque , I see that the "Answer" does contain some Sentences relevant to this Question. Point remains that the Sentences are hidden & this Question can be answered easily. – Prem Dec 16 '23 at 19:16

1 Answers1

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In modular arithmetic, especially when dealing with moduli like 8, negative numbers can indeed be solutions. However, in the context of modular arithmetic, all these solutions are considered equivalent to their positive counterparts. This is because modular arithmetic is concerned with the remainder after division, and both a positive integer and its negative counterpart will have the same remainder when divided by the modulus.

For example, in the case of $x^2 \equiv 1 \mod 8$:

  • $1^2 = 1 \equiv 1 \mod 8$
  • $3^2 = 9 \equiv 1 \mod 8$
  • $5^2 = 25 \equiv 1 \mod 8$
  • $7^2 = 49 \equiv 1 \mod 8$

Now, consider the negatives:

  • $(-1)^2 = 1 \equiv 1 \mod 8$
  • $(-3)^2 = 9 \equiv 1 \mod 8$
  • $(-5)^2 = 25 \equiv 1 \mod 8$
  • $(-7)^2 = 49 \equiv 1 \mod 8$

In each pair ($1$ and $-1$, $3$ and $-3$, etc.), the positive and negative numbers are congruent to each other modulo 8. This means they are considered the same in the context of modular arithmetic for mod 8. So, when you list the solutions, you don't need to list both the positive and negative versions separately; choosing either the positive or the negative ones suffices.

Ameer786
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  • Why don't we take the case with mod(5) and mod(7)? – FriedSpies Dec 10 '23 at 08:11
  • In modular arithmetic, negative solutions are equivalent to their positive counterparts, yielding the same remainder. Thus, for $x^2 \equiv 1$ modulo any number, including 5, 8, and 17, listing only the positive solutions is sufficient, as they implicitly include the negative ones. – Ameer786 Dec 10 '23 at 08:15
  • So, we don't have to count negatives in any case. This would lead to only 4 solutions. But there are 16 solutions indeed. – FriedSpies Dec 10 '23 at 08:19
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    You seem to claim that $1\equiv-1\bmod8,3\equiv-3\bmod8$, etc., which risks to confuse the OP. – Anne Bauval Dec 10 '23 at 11:59