Section 17.3 Using Eisenstein's Criterion
Let's calculate for a bit using this criterion. It says that we can tell whether a numberQuestion 17.3.1.
When does
Example 17.3.2.
Let's try with
so Theorem 17.2.8 asks for
What about with
This is even, and we already saw several times that this correctly implies 3 is a QR.
Claim 17.3.3.
Both of the parities we are adding can be easily computed:
The parity of
The parity of the size of the set of integers
such that
The sum of these two parities should be the parity of the set between
Proof.
We will actually compute both parities directly. The parity of \(k\) has two options.
If \(k\) is even, then \(k=2\ell\) and \(p=6k+r=12\ell+r\text{.}\)
If not, then \(k=2\ell+1\) and \(p=6k+r=12\ell+6+r\text{.}\)
To compute the second part, we first note that for prime \(p\text{,}\) the only possible residues \(r\) modulo \(6\) are \(r=1\) or \(r=5\text{.}\)
If \(r=1\text{,}\) we are looking for \(y\) such that \(\frac{1}{6}<y<\frac{1}{3}\text{,}\) of which there are none.
If \(r=5\text{,}\) we are looking for \(y\) such that \(\frac{5}{6}<y<\frac{5}{3}\text{,}\) of which there is one.
Proposition 17.3.4.
Three is a quadratic residue (or not) in the following circumstances.
if (mod ) if (mod )
Proof.
Combine the facts in Claim 17.3.3. We see that
If \(p=12\ell+1\) we add two even numbers, so 3 is a QR.
If \(p=12\ell+5\text{,}\) we add an even number and 1, so 3 is not a QR.
If \(p=12\ell+6+1=12\ell+7\text{,}\) we add an odd and zero, so 3 is not a QR.
If \(p=12\ell+6+5=12\ell +11\text{,}\) we add an odd and 1, which is even, so 3 is a QR.
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def _(p=prime_range(5,50)):
L = solve_mod(x^2==3,p)
pretty_print(html(r"$%s\equiv %s\text{ (mod }12)$ and $\left(\frac{3}{%s}\right)=%s$"%(p,p%12,p, legendre_symbol(3,p))))
if L:
pretty_print(html(r"And it turns out $%s^2\equiv %s$, $%s^2\equiv %s$ (mod $%s$)"%(L[0][0],L[0][0]^2,L[1][0],L[0][0]^2,p)))