Mersenne primes and the area of primitive pythagorean triangles

A triple \left ( a,b,c \right ) of positive integers, with a,b<c, is called pythagorean triple; that is  the lengths of the sides of pythagorean triangles, if and only if

a^{2}+b^{2}= c^{2}.

A pythagorean triangles is primitive if a, b and c are relatively prime, so that :

a= m^{2}-n^{2},~~~b= 2mn,~~~c= m^{2}+n^{2}

There is a method of finding primitive pythagorean triangles using Gaussian integers \mathbb{Z}[i]. [2]

Let \displaystyle a^2+b^2=c^2 where a,\,b, and c are pairwise, relatively prime, positive integers. Let z=a+bi and \bar{z}=a-bi, where i=\sqrt{-1}. Then z and \bar{z} are Gaussian Integers, and \bar{z} is the conjugate of z. Note that

\displaystyle a=\frac{z+\bar{z}}{2},\quad b=\frac{z-\bar{z}}{2i}\,,   and \displaystyle \quad c=\sqrt{z\bar{z}}.

Since \gcd(a,b)=1 then \gcd(z,\bar{z})=1. Hence each of z and \bar{z} is a square. That is, there exists integers m and n such that (m+ni)^2=z and (m-ni)^2=\bar{z}. So,

\displaystyle a =\frac{(m+ni)^2+(m-ni)^2}{2}=m^2-n^2,

\displaystyle b =\frac{(m+ni)^2-(m-ni)^2}{2i}=2mn,

\displaystyle \quad c =\sqrt{(m+ni)^2(m-ni)^2}=m^2+n^2.

Since a is a positive integer, m and n must be positive integers, m > n. And since \gcd(a,b)=1, m and n must be relatively prime and of opposite parity. We have the identity,

\displaystyle \left\vert \frac{z^{2k}+\bar{z}^{2k}}{2}\right\vert^2+\left\vert \frac{z^{2k}-\bar{z}^{2k}}{2i}\right\vert^2 =\Bigl(\left(z\bar{z}\right)^k\Bigr)^2.

The absolute values are necessary since the terms on the left, depending on k, may, or may not, be positive.

When the hypotenuse is to a power of the form c^{2^j}, where j is a non-negative integers, then there exists positive integers u and v such that :

\displaystyle u=\left\vert\frac{z^{2^j}+\bar{z}^{2^j}}{2}\right\vert=\left\vert\frac{(u+vi)+(u-vi)}{2}\right\vert, and \displaystyle \quad v=\left\vert\frac{z^{2^j}-\bar{z}^{2^j}}{2i}\right\vert=\left\vert\frac{(u+vi)-(u-vi)}{2i}\right\vert.

Note that, if u < v then interchange the labels. Clearly, \gcd(m,n)=1 implies \gcd(u,v)=1. The parameters m and n have opposite parity. Therefore 2\nmid \left(m^2+n^2\right)^{2^j}=u^2+v^2. Hence u and v have opposite parity.

Thus, all primitive Pythagorean triples of the form \left(a,b,c^{2^j}\right) are given by  the parametric equations :

\displaystyle a=u^2-v^2=\left\vert\frac{(m+ni)^{2^{j+1}}+(m-ni)^{2^{j+1}}}{2}\right\vert,

\displaystyle b=2uv=\left\vert\frac{(m+ni)^{2^{j+1}}-(m-ni)^{2^{j+1}}}{2i}\right\vert,

and \displaystyle c^{2^j}=u^2+v^2=\left(m^2+n^2\right)^{2^j}.

Then

\displaystyle 2ab=\left\vert\frac{(m+ni)^{2^{j+2}}-(m-ni)^{2^{j+2}}}{2i}\right\vert.

Barnes[1] showed that if \left(a,b,c^{2^j}\right) is a solution to a primitive Pythagorean triangle, where j is a non-negative integer, then every Mersenne prime less than or equal to \left(2^{j+2}-1\right) divides ab. This imples that the area of the primitive pythagorean triangle is a multiple of the perfect number.

Let t be a positive integer such that 2^{t+1}-1 is a prime (Mersenne prime). And let a^2+b^2=c^{2^t} be a primitive Pythagorean triangle. Then, the area \frac{1}{2}\,ab is a multiple of the perfect number 2^t\left(2^{t+1}-1\right).

Theorem 1

Let a^2+b^2=\left(c^k\right)^2 be a primitive Pythagorean triangle where k \in \mathbb{Z}^+. Let d be a positive integer divisor of k. Let p=4d-1. If p is a prime then p\vert ab.

Proof

Since m+ni\in \mathbb{Z}[i], then

\left((m+ni)^{\frac{k}{d}}\right)^{4d}=(M+Ni)^{4d}

for some M,N\in \mathbb{Z}.

If p=4d-1 is a prime, then since -i=i^{4d-1}=i^p,

\displaystyle p\vert(M+Ni)^p-(M-Ni)=\bigl((M+Ni)^p-M^p-(Ni)^p\bigr)+(M^p-M)-i(N^p-N).

Which implies

\displaystyle p\vert(M+Ni)^{p+1}-\left(M^2+N^2\right).

Similarly,

\displaystyle p\vert(M-Ni)^{p+1}-\left(M^2+N^2\right).

Which implies

\displaystyle p\,\, divides \displaystyle \,\,(M+Ni)^{p+1}-(M-Ni)^{p+1}=(m+ni)^{4k}-(m-ni)^{4k}=4abi.

Therefore p\vert ab.

Corollary 1

Let a^2+b^2=\left(c^{2^j}\right)^2 be a primitive Pythagorean triangle where j is a nonnegative integer. Let M be any Mersenne prime less than or equal to 2^{j+2}-1. Then

M divides ab.

Proof

Let k=2^j in theorem (1). Then

\displaystyle d=\left\{\,2^0,\,2^1,\,2^2,\ldots ,2^{j-1},\,2^j\,\right\},
\displaystyle 4d-1=\left\{\,2^2-1,\,2^3-1,\,2^4-1,\ldots ,2^{j+1}-1,\,2^{j+2}-1\,\right\},

From Theorem (1), every prime in the set 4d-1 divides ab.

Theorem 2

If

\displaystyle a^2+b^2=\left(c^{2^j}\right)^2

is a primitive Pythagorean triangle where j is a nonnegative integer then 2^{j+2} divides ab.

Proof

By induction on j : If j=0 then \displaystyle a^2+b^2=\left(c^{2^0}\right)^2=c^2. So there exists integers m and n, one odd the other even, such that a=m^2-n^2 and b=2mn. Hence 4=2^{0+2} divides ab. Assume true for j, then

\displaystyle \left(c^{2^{j+1}}\right)^2=\left[\left(c^{2^j}\right)^2\right]^2=\left(a^2+b^2\right)^2=\left(a^2-b^2\right)^2+(2ab)^2

Let a_1=a^2-b^2 and b_1=2ab. Then if 2^{j+2} divides ab, 2^{(j+1)+2} divides a_1b_1.

Perfect numbers and the area of primitive pythagorean triangles :

From corollary and theorem 2 we see that the area of the 3\--4\--5 triangle, the smallest Pythagorean triangle, is \frac{1}{2}\left(2^2-1\right)\left(2^2\right)=2^1\left(2^2-1\right)=6, the smallest perfect number.

Let t be a positive integer such that 2^{t+1}-1 is a prime (Mersenne prime). And let a^2+b^2=c^{2^t} be a primitive Pythagorean triangle. Then, the area \frac{1}{2}\,ab is a multiple of the perfect number 2^t\left(2^{t+1}-1\right).

Examples
t=1 : Two primitive Pythagorean triangles 3^2+4^2=5^{2^1} and 5^2+12^2=13^{2^1}. Since 2^{1+1}-1=2^2-1=3 is a prime, we know that each area must be a multiple of the perfect number 2^1\left(2^2-1\right)=6. And indeed, \frac{1}{2}\,(3)(4)=\mathbf{6} and \frac{1}{2}\,(5)(12)=\mathbf{6}\,(5).

t=2 : Two primitive Pythagorean triangles 7^2+24^2=5^{2^2} and 119^2+120^2=13^{2^2}. Since 2^{2+1}-1=2^3-1=7 is a prime, we know that each area must be a multiple of the perfect number 2^2\left(2^3-1\right)=28. And indeed, \frac{1}{2}\,(7)(24)=\mathbf{28}\,(3) and \frac{1}{2}\,(119)(120)=\mathbf{28}\,(255).

t=4 : Two primitive Pythagorean triangles 164833^2+354144^2=5^{2^4} and 815616479^2+13651680^2=13^{2^4}. Since 2^5-1=31 is a prime, we know that each area must be a multiple of the perfect number 2^4\left(2^5-1\right)=496. And indeed, \frac{1}{2}\,(164833)(354144)=\mathbf{496}\,(58845381) and \frac{1}{2}\, (815616479)(13651680)=\mathbf{496}\,(11224329812535).

Note: Since 5^{2^4}=\left(5^8\right)^{2^1}=\left(5^4\right)^{2^2}, \frac{1}{2}\,(164833)(354144) is also a multiple of each of 6 and 28.

That is,

  • If a^2+b^2=c^2 is a primitive Pythagorean triangle then its area is a multiple of the perfect number \mathbf{6}.
  • If a^2+b^2=c^4 is a primitive Pythagorean triangle then its area is a multiple of each of the perfect numbers \mathbf{6} and \mathbf{28}
  • If a^2+b^2=c^{16} is a primitive Pythagorean triangle then its area is a multiple of each of the perfect numbers \mathbf{6},\,\mathbf{28}, and \mathbf{496}
  • If a^2+b^2=c^{64} is a primitive Pythagorean triangle then its area is a multiple of each of the perfect numbers \mathbf{6},\,\mathbf{28},\,\mathbf{496}, and \mathbf{8128}.
  • If a^2+b^2=c^{4096} is a primitive Pythagorean triangle then its area is a multiple of each of the perfect numbers \mathbf{6},\,\mathbf{28},\,\mathbf{496},\,\mathbf{8128}, and \mathbf{33550336}
  • And, in general, if a^2+b^2=\left(c^{2^t}\right) is a primitive Pythagorean triangle where 2^{t+1}-1 is a Mersenne prime then the triangle’s area is a multiple of each perfect number less than or equal to the perfect number 2^t\left(2^{t+1}-1\right).

References :

[1] Barnes, Fred. Primitive Pythagorean triangles where the hypotenuse is to a power, available at pythag.net/node6.html

[2] Conrad, Keith. The Gaussian Integers, available at math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf

One thought on “Mersenne primes and the area of primitive pythagorean triangles

  1. Very interesting. I noticed they found a new Mersenne prime today, started wondering if there was a way to predict Mersenne primes based on Pythagorean triples. I see you have already been thinking along these lines. Thanks.

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