Section7.5Epilogue: Why congruences matter
Before we move to greater abstraction, let's take a final look at whether you can find a pattern in the primes \(p\) such that \[x^2\equiv -1\text{ mod }(p)\] has a solution! Take a look at this table and see if you can find something.
The reason I point this kind of thing out is not just because I can, but because it shows simple congruence patterns can have a big result. Here's another example.
Recall our search through Mordell/Bachet curves, and let's look at the particular case \(y^2=x^3+7\).
It's amazing how the curve slips between every integer lattice point... So it seems that a perfect square can't ever be exactly seven more than a perfect cube. Is this true? Here's where congruences come into play.
Proposition7.5.1Showing a Mordell curve has no integer point
There is no integer \(x\) such that \(y^2=x^3+7\), so there are no integer points on this curve.Proof
First let's consider the case where \(x\) is even. Then \(2\mid x\), so \(8\mid x^3\). 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 \(4\), so that's not possible.
- So it must be the case that \(x\equiv 1\) mod \((4)\).
Now we do a trick like that of completing the square: \[y^2=x^3+7\Rightarrow y^2+1=x^3+8\Rightarrow y^2+1= (x+2)(x^2-2x+4)\] 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 of the form \(4n+3\).
- (Otherwise all its primes look like \(4n+1\equiv 1\), the product of which would also be \(\equiv 1\) mod \((4)\).)
- If such a prime divides \(x+2\), it (naturally) divides \((x+2)(x^2-2x+4)\) as well.
- But if it divides \((x+2)(x^2-2x+4)\), it must then divide \(y^2+1\), since they're equal.
- However, our exploration (see exercises) said that a prime of the form \(4n+3\) can't divide \(y^2+1\)! So this option doesn't work either.
Enough said; congruences are amazingly powerful.
