In 'Elements of Number Theory' John Stillwell considers the division property in the integer-like sets of numbers Z[i] (the Gaussian integers, p107) and Z[√-2] (an example of a set of quadratic integers, p119). In the case of the Gaussian integers it is relatively easy to visualize 'multiples' of integers as forming a square grid as in Figure 6.1. However, for the quadratic integers Z[√-2] this doesn't appear to be so easy and you wonder how the grid in Figure 7.1 arises and how Pythagoras' theorem in this case can be applied in the proof.
I will attempt to go through both cases and explain how I see it. Firstly, the Gaussian integers. Suppose μ and β are the Gaussian integers
$$ \mu=m+ni...(1)$$
$$ \beta=r+si...(2)$$
where m, n, r and s are elements of Z and i=√-1. Then
$$\mu\beta=m\beta+n\beta i...(3)$$
Now we have that
$$\beta i=ri+si^{2}=-s+ri...(4)$$
If we think of β and βi as vectors on the complex plane, then β is the vector (r,s) and βi is the vector (-s,r). So, as Stillwell says, i rotates the vector β anticlockwise through 90 degrees. We can see that using the standard dot product of these vectors
$$(r,s)\cdot(-s,r)=r(-s)+sr=0...(5)$$
as expected for mutually orthogonal vectors. We can see therefore that equation (3) is telling us that μβ is the sum of real multiples of β and βi. We can therefore think of this as a square grid with β and βi as base vectors, as anticipated.
Now what about the quadratic integers Z[√-2]? Suppose that this time μ and β are the quadratic integers
$$ \mu=a+b\sqrt{-2}...(6)$$
$$ \beta=c+d\sqrt{-2}...(7)$$
where a, b, c, and d are elements of Z. Then
$$\mu\beta=a\beta+b\beta\sqrt{-2}...(8)$$
What is β√-2? We have
$$\beta\sqrt{-2}=c\sqrt{-2}+d(\sqrt{-2})^{2}=-2d+c\sqrt{-2}...(9)$$
If we again think about μ and β as being vectors, then β is the vector (c,d) and β√-2 is the vector (-2d,c). In the normal sense it does not look like these two vectors are orthogonal, however, if we define the dot product of two vectors (w,x) and (y,z) to be
$$(w,x)\cdot(y,z)=wy+2xz...(10)$$
then
$$(c,d)\cdot(-2d,c)=-2cd+2dc=0...(11)$$
and so we retrieve the orthogonality of β and β√-2. The definition in equation (10) also fits in with the definition of the norm since
$$norm\beta=(c,d)\cdot(c,d)=c^{2}+2d^{2}...(12)$$
which is as defined by Stillwell (p120). Hence we can again think that equation (8) is telling us that μβ is the sum of real multiples of β and β√-2; i.e. we can think of a grid with β and β√-2 being base vectors, as described in the book (Figure 7.1). Note, however, that this is not a square grid as β and β√-2 are not the same length.
I want to go one step further and show that the proof on page 119 doesn't have to rely on Pythagoras' Theorem. We can see from Figure 7.1 that |ρ| is less than or equal to the length of the vector from the origin to the centre of the figure. That is the vector
$$\frac{\beta}{2}+\frac{\beta\sqrt{-2}}{2}...(13)$$
which is
$$\frac{1}{2}(c,d)+\frac{1}{2}(-2d,c)=\frac{1}{2}(c-2d,d+c)...(14)$$
It follows that
$$\left|\rho\right|^{2}\le\frac{1}{2}(c-2d,d+c)\cdot\frac{1}{2}(c-2d,d+c)...(15)$$
Expanding the RHS we get
$$ \frac{1}{2}(c-2d,d+c)\cdot\frac{1}{2}(c-2d,d+c)=\frac{1}{4}((c-2d)^{2}+2(d+c)^{2})=\frac{1}{4}(c^{2}-4cd+4d^{2}+2d^{2}+4cd+2c^{2})...(16)$$
and so
$$
\frac{1}{2}(c-2d,d+c)\cdot\frac{1}{2}(c-2d,d+c)=\frac{1}{4}(3c^{2}+6d^{2})=\frac{3}{4}(c^{2}+2d^{2})=\frac{3}{4}\left|\beta\right|^{2}...(17)$$
Hence we have
$$\left|\rho\right|^{2}\le\frac{3}{4}\left|\beta\right|^{2}...(18)$$
as obtained in the book.
Friday 23 January 2015
Sunday 4 January 2015
Geometric characterization of the primes that are sums of two squares
Previously, we saw that Fermat's two square theorem says that all primes of the form 4n+1 are the sum of two integer squares. Using Euclid's integer solutions to the Pythagorean equation z2=x2+y2 we obtain the following geometric characterization of the primes that are the sum of two squares (Elements of Number Theory p112):-
"The primes that are the sums of two squares are those that occur as hypotenuses of right-angled triangles with integer sides."
This is a nice result. We can now imagine all of these sorts of primes as being part of right-angled triangles with integer sides and there are infinitely many of them. It is a very satisfactory geometric picture of these types of primes. However, is the triangle with hypotenuse prime p unique? This is the basis of Exercise 6.6.2 (p112):-
"6.6.2. Given a prime p=4n+1, is the integer right-angled triangle with hypotenuse p unique?"
As an example of some of the work that I have done with this book, I thought I would show how I proved this.
We aim for a proof by contradiction. Given a prime p=4n+1 we suppose to the contrary that the integer right-angled triangle with hypotenuse p is not unique. So there are at least two integer right-angled triangles with hypotenuse p. Let the non-hypotenuse sides of two of these triangles be a1, b1, and a2, b2.
Firstly, we can show that a1 and b1 are coprime. Suppose that they are not then there exists a positive integer c (not equal to 1) that divides a1 and b1 such that a1=cx1 and b1=cy1. Then by Pythagoras for a right-angled triangle:-
$$ a_{1}^{2}+b_{1}^{2}=p^{2} $$
$$ \Rightarrow c^{2}x_{1}^{2}+c^{2}y_{1}^{2}=p^{2}$$
$$ \Rightarrow c^{2}(x_{1}^{2}+y_{1}^{2})=p^{2}$$
Now as c divides the LHS, c must divide p2, but the only divisors of p2 are 1, p or p2. If c was p2 this would imply
$$ x_{1}^{2}+y_{1}^{2}=1/p^{2}$$
and x1 and y1 would not be integers. If c was p then this would imply
$$ x_{1}^{2}+y_{1}^{2}=1$$
and so either x1=0 and y1=1 or x1=1 and y1=0 and these are not right-angled triangles. Hence c=1 and we have reached a contradiction. Thus a1 and b1 are coprime (as are a2 and b2).
Hence, by the theorem for Primitive Pythagorean Triples (top of page 112) we can say that
$$a_{1}=u_{1}^{2}-v_{1}^{2}$$
$$b_{1}=2u_{1}v_{1}$$
$$a_{2}=u_{2}^{2}-v_{2}^{2}$$
$$b_{2}=2u_{1}v_{2}$$
for some integers u1,v1,u2,v2. It follows that
$$a_{1}^{2}+b_{1}^{2}=(u_{1}^{2}+v_{1}^{2})^{2}=p^{2}$$
and
$$a_{2}^{2}+b_{2}^{2}=(u_{2}^{2}+v_{2}^{2})^{2}=p^{2}$$
However, we know that as p=4n+1 there is only one u, v such that p=u2+v2 (top of page 110) and so there is only one u, v such that p2=(u2+v2)2. Thus, we have reached a contradiction. Hence the integer right-angled triangle with hypotenuse p must be unique.
"The primes that are the sums of two squares are those that occur as hypotenuses of right-angled triangles with integer sides."
This is a nice result. We can now imagine all of these sorts of primes as being part of right-angled triangles with integer sides and there are infinitely many of them. It is a very satisfactory geometric picture of these types of primes. However, is the triangle with hypotenuse prime p unique? This is the basis of Exercise 6.6.2 (p112):-
"6.6.2. Given a prime p=4n+1, is the integer right-angled triangle with hypotenuse p unique?"
As an example of some of the work that I have done with this book, I thought I would show how I proved this.
We aim for a proof by contradiction. Given a prime p=4n+1 we suppose to the contrary that the integer right-angled triangle with hypotenuse p is not unique. So there are at least two integer right-angled triangles with hypotenuse p. Let the non-hypotenuse sides of two of these triangles be a1, b1, and a2, b2.
Firstly, we can show that a1 and b1 are coprime. Suppose that they are not then there exists a positive integer c (not equal to 1) that divides a1 and b1 such that a1=cx1 and b1=cy1. Then by Pythagoras for a right-angled triangle:-
$$ a_{1}^{2}+b_{1}^{2}=p^{2} $$
$$ \Rightarrow c^{2}x_{1}^{2}+c^{2}y_{1}^{2}=p^{2}$$
$$ \Rightarrow c^{2}(x_{1}^{2}+y_{1}^{2})=p^{2}$$
Now as c divides the LHS, c must divide p2, but the only divisors of p2 are 1, p or p2. If c was p2 this would imply
$$ x_{1}^{2}+y_{1}^{2}=1/p^{2}$$
and x1 and y1 would not be integers. If c was p then this would imply
$$ x_{1}^{2}+y_{1}^{2}=1$$
and so either x1=0 and y1=1 or x1=1 and y1=0 and these are not right-angled triangles. Hence c=1 and we have reached a contradiction. Thus a1 and b1 are coprime (as are a2 and b2).
Hence, by the theorem for Primitive Pythagorean Triples (top of page 112) we can say that
$$a_{1}=u_{1}^{2}-v_{1}^{2}$$
$$b_{1}=2u_{1}v_{1}$$
$$a_{2}=u_{2}^{2}-v_{2}^{2}$$
$$b_{2}=2u_{1}v_{2}$$
for some integers u1,v1,u2,v2. It follows that
$$a_{1}^{2}+b_{1}^{2}=(u_{1}^{2}+v_{1}^{2})^{2}=p^{2}$$
and
$$a_{2}^{2}+b_{2}^{2}=(u_{2}^{2}+v_{2}^{2})^{2}=p^{2}$$
However, we know that as p=4n+1 there is only one u, v such that p=u2+v2 (top of page 110) and so there is only one u, v such that p2=(u2+v2)2. Thus, we have reached a contradiction. Hence the integer right-angled triangle with hypotenuse p must be unique.
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