1
$\begingroup$

To determine a quadratic congruence equation has any solution, we have to evaluate $$\bigg( \dfrac{a}{p} \bigg)$$, So can we apply the algorithm of Jacobi symbol to evaluate this? If yes, what's the easiest way to evaluate this symbol? Let's not take efficiency into account because I don't have a computer in my exam, so I just want to learn the simplest way which I will be able recall it easily during the exam. For example, evaluate $$\bigg( \dfrac{2819}{4177} \bigg)$$

$$(\dfrac{2819}{4177}) = (\dfrac{4177}{2819}) \text{ because } \frac{(2819-1)(4177-1)}{4} = \text{ even } $$ $$= (\dfrac{1358}{2819}) = (\dfrac{2}{2819}) \cdot (\dfrac{679}{2819})$$ $$= (-1) \cdot (\dfrac{679}{2819}) \text{ because } \frac{2819^2 - 1}{8} = \text{ odd }$$ $$= (-1) \cdot (-1) \cdot (\dfrac{2819}{679}) = (\dfrac{2819}{679}) \text{ because } \frac{(m-1)(n-1)}{4} = \text{ odd }$$ $$= (\dfrac{2819}{679}) = (\dfrac{103}{679}) = (-1) \cdot (\dfrac{679}{103}) \text{ because } \frac{(103-1)(679-1)}{4} = \text{ odd }$$ $$= -(\dfrac{61}{103}) = -(\dfrac{103}{61}) \text{ because } \frac{(61-1)(103-1)}{4} = \text{ even }$$ $$= -(\dfrac{42}{61}) = -(\dfrac{2}{61}) \cdot (\dfrac{21}{61}) = -(\dfrac{21}{61}) \text{ because } \frac{61^2 - 1}{8} = \text{ even }$$ $$= -(\dfrac{61}{21}) \text{ because } \frac{(21-1)(61-1)}{4} = \text{ even }$$ $$= -(\dfrac{19}{21}) = -(\dfrac{21}{19}) \text{ because } \frac{(19-1)(21-1)}{4} = \text{ even }$$ $$= -(\dfrac{2}{19}) = (-1) \cdot (-1) = 1 \text{ because } \frac{19^2-1}{8} = \text{ odd }$$.

Since in the book example, the author often skipped many steps, so I just want to make sure that I understand it correctly. Any suggestion and idea would be greatly appreciated.

Thank you,

$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

It looks to me like what you have done is fine. Did you check to see whether 4177 is prime? If it isn't, then Jacobi symbol 1 doesn't guarantee quadratic congruence solvable, e.g., $x^2\equiv2\pmod{15}$ has no solution even though the symbol is 1.

$\endgroup$
3
  • $\begingroup$ Thanks. 4177 is actually prime. So like you said, before attempting to evaluate Jacobi symbol, I have to check if $p$ is prime? $\endgroup$
    – roxrook
    Apr 16, 2011 at 6:52
  • 1
    $\begingroup$ @Chan: You can evaluate Jacobi symbols $\left(\frac{a}{m}\right)$ without caring about whether $m$ is prime. But then, as pointed out in the comment above, if the result is $1$ you cannot conclude that $a$ is a quadratic residue of $m$. (If $m$ is prime, of course you can.) If the Jacobi symbol evaluates to $-1$, however, you can conclude that $a$ is not a quadratic residue of $m$, whether or not $m$ is prime. $\endgroup$
    – André Nicolas
    Apr 16, 2011 at 12:49
  • $\begingroup$ @user6312: Thanks a lot. $\endgroup$
    – roxrook
    Apr 16, 2011 at 17:56
1
$\begingroup$

There are two problems. First, you have an error above: $\:(2|61) = -1\:,\:$ not $1$. Second, as I mentioned before when discussing the generalized Euler Criterion:

BEWARE $\ $ The criterion cannot be expressed equivalently as a simple Jacobi symbol calculation. For example we have $\rm(8|15) = 1\ $ but $8$ is not a square (mod $15$).

However, in your example all is well since, after correcting the sign error, the Jacobi symbol evaluates to $-1$ so it correctly identifies nonsquares. But when it evaluates to $1$ (as you had above) it needn't correctly identify squares (except when the denominator is prime, i.e. when it reduces to a Legendre symbol).

$\endgroup$
5
  • $\begingroup$ Thank you Bill, so what is the main difference between Jacobi and Legendre? I'm a bit confused now :(. $\endgroup$
    – roxrook
    Apr 16, 2011 at 17:56
  • 1
    $\begingroup$ @Chan, for $(p|q)$ to be a Legendre symbol, $q$ must be prime. A Legendre symbol is 1 if $p$ is a square mod $q$, $-1$ if it isn't. For $(p|q)$ to be a Jacobi symbol, there is no such restriction on $q$. If the symbol is $-1$, $p$ isn't a square, but if it's 1, you can't say whether $p$ is a square. $\endgroup$
    – Gerry Myerson
    Apr 16, 2011 at 23:58
  • $\begingroup$ @Gerry Myerson: Many thanks. So if I want to see if is there a solution for quadratic congruence, then I should stick with Legendre symbol? $\endgroup$
    – roxrook
    Apr 17, 2011 at 4:48
  • 1
    $\begingroup$ @Chan: You can use the generalized Euler criterion - see the link above (originally the link pointed to the wrong place - I just fixed it). $\endgroup$
    – Bill Dubuque
    Apr 17, 2011 at 7:04
  • $\begingroup$ Many thanks, now I found it much easier ;) $\endgroup$
    – roxrook
    Apr 18, 2011 at 4:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .