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.