Section 7.6 Epilogue: Why Congruences Matter
Although we will spend some significant time working on solving congruences, I don't want to lose sight of deeper questions. To see how congruences help address them, recall the search in Section 7.1 for primesxxxxxxxxxx
import itertools
β
def _(n=20):
yeslist=[]
nolist=[]
for p in prime_range(3,n):
res = 0
for res in [0..p]:
if mod(res,p)^2+1 == 0:
yeslist.append(p)
break
else:
nolist.append(p)
t = [['exist', 'do not exist']] + [[a,b] for (a,b) in itertools.zip_longest(yeslist,nolist)]
for item in t:
for i in range(len(item)):
if item[i] is None:
item[i]=''
pretty_print(html(r"Solutions to $x^2\equiv -1$ (mod $p$) for $2\le p \le %s$:"%n))
pretty_print(html(table(t, header_row = True, frame = True)))
Question 7.6.1.
Do you see a pattern related to some kind of congruence? (This one should be more apparent than in Section 7.3; see also Exercise 7.7.12.)

Proposition 7.6.3. Showing a Mordell curve has no integer point.
There are no integers
Proof of Proposition 7.6.3.
For convenience, we will rewrite \(x^3=y^2-7\) as \(y^2=x^3+7\text{.}\) To begin the proof, first consider the case where \(x\) is even. Then \(2\mid x\text{,}\) so \(8\mid x^3\text{.}\) That means \(y^2\equiv 7\) (mod \(8\)).
Unfortunately, the only perfect squares mod \((8)\) seem to be 0, 1, and 4. So this is not possible.
What about if \(x\) is odd? Then \(y\) must be even, since \(x^3\) and \(7\) are odd. So let's examine whether \(x\equiv 1\) (mod \(4\)) or \(x\equiv 3\) (mod \(4\)), the next two options.
If \(x\equiv 3\) (mod \(4\)), then \(x^3\equiv 27\equiv 3\) (mod \(4\)), so \(x^3+7\equiv 10\equiv 2\) (mod \(4\)). But we already know from earlier that perfect squares are only \(0\) or \(1\) modulo \(4\text{,}\) so that's not possible.
So it must be the case that \(x\equiv 1\) (mod \(4\)).
Now we do a trickβ5β like that of completing the square:
Let's analyze this carefully in the following argument.
If \(x\equiv 1\) (mod \(4\)), then \(x+2\equiv 3\) (mod \(4\)).
So not only is \(x+2\) an odd number, but also it must be divisible by a prime \(q\) of the form \(4n+3\text{.}\) (Otherwise all its primes look like \(4n+1\equiv 1\text{,}\) the product of which would also be \(\equiv 1\) (mod \(4\)).)
If \(q\) divides \(x+2\text{,}\) it (naturally) divides \((x+2)(x^2-2x+4)\) as well. But if it divides \((x+2)(x^2-2x+4)\text{,}\) it must then divide \(y^2+1\text{,}\) since they're equal.
However, our exploration at the start of this section suggested that a prime of the form \(4n+3\) can't divide \(y^2+1\text{!}\)
So, assuming it is true that only primes of the form \(4n+1\) can divide perfect squares plus one (\(y^2+1\)), then \(x\equiv 1\text{ (mod }4)\) doesn't work either.