prove that if $a = bmod(p)$ then the value of the legendre symbol $(\frac{a}p) = (\frac{b}p)$

Question

I started by using the fact that $a = b + \lambda p$ and i then substituted this into eulers lemma: $a^{(p-1)/2} = (\frac{b + \lambda p}p)mod(p)$. But i was unsure if you could split up the the symbol $(\frac{b + \lambda p}p)$ into $(\frac{b}p) + (\frac{\lambda p}p)$. because if so then the proof follows quite quickly. but if not i dont know how to proceede.

Answer 1

All we need is the definition of the Legendre aymbol. Suppose that $a\equiv b\pmod{p}$. Then the congruence $x^2\equiv a\pmod{p}$ has a solution if and only if the congruence $x^2\equiv b\pmod{p}$ has a solution.

Thus if $(a/p)=1$ then $(b/p)=1$ and if $(a/p)=-1$ then $(b/p)=-1$.

Answer 2

Your approach indeed works (though it's an overkill ) . Use Euler's lemma :

$$\left (\frac{a}{p} \right )\equiv a^{\frac{p-1}{2}} \pmod{p}$$

$$ \left (\frac{b}{p} \right )\equiv b^{\frac{p-1}{2}} \pmod{p}$$

Now because $a \equiv b \pmod{p}$ it follows that $$a^{\frac{p-1}{2}} \equiv b^{\frac{p-1}{2}} \pmod{p}$$

Now use all these congruences together so :

$$\left (\frac{a}{p} \right )\equiv \left (\frac{b}{p} \right ) \pmod{p}$$

But remember that these numbers are $-1,0$ or $1$ so because $p>2$ they must be equal else the congruence doesn't hold .