# A contradiction in calculating the legendre symbol

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

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

But I can't find the mistakes.

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

• Please do not delete after having received an answer. – 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 $$\left(\frac{2}{13}\right)=(-1)^{21}=-1.$$
• oops!! Yea, you are right. The denominator should be $8$. Thanks! – Sam Wong May 13 '19 at 13:53