I got a contradiction when I calculated the legendre symbol. I felt like there must be something wrong in my calculation but I can't find them. The following is my calculation steps.

Note that $13=4\times 3+1$, which is of the form $4n+1$.

$({2\over 13})=(-1)^k$, where $k=\frac{13^2-1}{2}=84.$ So $({2\over 13})=1.$ (The first equality is by a well-known formular.)

So $({8\over 13})=({2\over 13})\cdot({2\over 13})\cdot({2\over 13})=1.$

And, $({-1\over 13})=(-1)^m$, where $m=\frac{13-1}{2}=6.$ So $({-1\over 13})=1$.

Now let's calculate $({5\over 13})$.

$({5\over 13})=({13\over 5})=({3\over 5})=(-1)^t$, where $t=[\frac{3}{5}]+[\frac{6}{5}]=0+1=1$, where $f(x)=[x]$ is defined as taking the integer part of $x$. The last equality is also by a standard result.

So $({5\over 13})=-1$.

Here the contradiction comes.

$-1=({5\over 13})=({-8\over 13})=({-1\over 13})({8\over 13})=1\cdot1=1,$ which is a contradiction.

But I can't find the mistakes.

Can anyone help me check my calculation? Thanks a lot.

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    $\begingroup$ Please do not delete after having received an answer. $\endgroup$ – quid May 14 '19 at 15:27

Your first formula is wrong; in general you have $({2\over p})=(-1)^k$ where $k=\tfrac{p^2-1}{8}$.

In this particular case, with $p=13$, you get $k=\tfrac{13^2-1}{8}=21$ and so $$\left(\frac{2}{13}\right)=(-1)^{21}=-1.$$

  • $\begingroup$ oops!! Yea, you are right. The denominator should be $8$. Thanks! $\endgroup$ – Sam Wong May 13 '19 at 13:53
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    $\begingroup$ I am so stupid...Thanks again $\endgroup$ – Sam Wong May 13 '19 at 13:53
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    $\begingroup$ @SamWong: Don't call yourself stupid. Even if you make a stupid mistake, it doesn't imply that you're stupid. In fact, only stupid people don't admit their mistakes. =) $\endgroup$ – user21820 May 16 '19 at 7:12

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