Why is the Legendre symbol not defined for $p = 2$ (even prime) ?

According to the definition of the Legendre symbol $$\left(\frac a p\right)$$ it is defined for an odd prime $p$ only. Even thus the case where $p = 2$ is an even prime is not particularly interesting, why are the restriction made that $p$ must be an odd prime ?

Source: http://en.wikipedia.org/wiki/Legendre_symbol

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    $\begingroup$ The answer is mostly because Legendre's original formulation for the symbol was $a^{\frac{p-1}{2}}\pmod{p}, which isn't necessarily well-defined for the case of even p since it creates a fractional power. $\endgroup$ – Foo Barrigno Apr 1 '14 at 11:14
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    $\begingroup$ You mean we get $a^{\frac 1 2} = \sqrt{a}$ and we dont want to work with non-integers in number-theory ? According to the new definition, why not change the domain to include $p=2$ ? $\endgroup$ – Shuzheng Apr 1 '14 at 11:19
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    $\begingroup$ that's essentially what the Jacobi symbol is (en.wikipedia.org/wiki/Jacobi_symbol), but again it only applies to odd integers (not composites), and Kronecker symbol (en.wikipedia.org/wiki/Kronecker_symbol) generalizes it to all integers. $\endgroup$ – Foo Barrigno Apr 1 '14 at 11:23

There isn't a way to define it that helps to formulate the law of quadratic reciprocity, which is what Legendre was trying to do when introducing his symbol.

As I'm sure you observed, every $a$ is a quadratic residue mod 2, but 2 is not a quadratic residue modulo every modulus, or modulo every prime modulus. So computing when 2 is a quadratic residue mod $p$ is interesting (for odd primes $p$), but computing when $p$ is a quadratic residue mod 2 is trivial.

Added: If we wanted to define the Legendre symbol for $p=2$, just for the sake of completeness, the natural definition would be $(a/p) = 1$ for $a$ odd and $(a/p) = 0$ for $a$ even. It's important that we keep in mind that quadratic reciprocity only works with two odd primes, so omitting $p=2$ from the definition of the Legendre symbol is justifiable for avoiding any confusion on that point. What this does "make work" is the multiplicative character of the symbol, i.e. $(ab/p) = (a/p)(b/p)$ now also when $p=2$.


The Legendre symbol $(a/p)$ is a multiplicative character of the finite field of order $p$, $\mathbb{F}_{p}$. Since it maps onto $\pm 1$ it is of order $2$. A multiplicative character of $\mathbb{F}_{p}$ has to have order dividing $p-1$, so the Legendre symbol is only properly defined when $p$ is odd.

You can make an arbitrary definition for $p = 2$, but without the interesting properties of a multiplicative character, your definition is unlikely to be mathematically interesting.


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