- #1
Chris L T521
Gold Member
MHB
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Thanks again to those who participated in last week's POTW! Here's this week's problem!
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Problem: Let $p$ be an odd prime and $a\in\mathbb{Z}$. Define the Legendre symbol
\[\left(\frac{a}{p}\right)= \begin{cases}1 & \text{if $x^2\equiv a\pmod{p}$ has an integer solution} \\ 0 & \text{if $p\mid a$}\\ -1 & \text{if $x^2\equiv a\pmod{p}$ has no integer solution}\end{cases}\]
If $p$ is an odd prime and $a,b$ are two integers such that $(p,ab)=1$, then prove that
\[\left(\frac{ab}{p}\right)=\left(\frac{a}{p}\right) \left(\frac{b}{p}\right).\]
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Hint:
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Problem: Let $p$ be an odd prime and $a\in\mathbb{Z}$. Define the Legendre symbol
\[\left(\frac{a}{p}\right)= \begin{cases}1 & \text{if $x^2\equiv a\pmod{p}$ has an integer solution} \\ 0 & \text{if $p\mid a$}\\ -1 & \text{if $x^2\equiv a\pmod{p}$ has no integer solution}\end{cases}\]
If $p$ is an odd prime and $a,b$ are two integers such that $(p,ab)=1$, then prove that
\[\left(\frac{ab}{p}\right)=\left(\frac{a}{p}\right) \left(\frac{b}{p}\right).\]
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Hint:
One way to prove the identity is to use Euler's Criterion, which says the following:
Let $p$ be an odd prime and $a$ an integer such that $(a,p)=1$. Then
\[a^{\frac{p-1}{2}}\equiv \left(\frac{a}{p}\right)\pmod{p}.\]
Let $p$ be an odd prime and $a$ an integer such that $(a,p)=1$. Then
\[a^{\frac{p-1}{2}}\equiv \left(\frac{a}{p}\right)\pmod{p}.\]