Area of a primitive Pythagorean triangle and the perfect numbers

A perfect number is a positive integer that is equal to the sum of its divisors excluding itself. For example $6$ is a perfect number since its divisors are $1,\,2,\,3$ and $6$ where $1+2+3=6.$ All even perfect numbers are of the form $2^{s-1}\left(2^s-1\right)$ where $2^s-1$ is a prime. Such primes are called Mersenne primes. The first five perfect numbers are

1. $2^1\left(2^2-1\right)=6$
2. $2^2\left(2^3-1\right)=28$
3. $2^4\left(2^5-1\right)=496$
4. $2^6\left(2^7-1\right)=8128$
5. $2^{12}\left(2^13-1\right)=33550336$

It's not known if there are any odd perfect numbers.

From corollary (1) and theorem (2) we see that it is no coincidence 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, as table (4) shows, the area $\frac{1}{2}\,ab$ is a multiple of the perfect number $2^t\left(2^{t+1}-1\right).$

Examples

Consider the two primitive Pythagorean triangles $3^2+4^2=5^{2^1}$ and $5^2+12^2=13^{2^1}.$ In this case $t=1.$

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 $\frac{1}{2}\,(5)(12)=\mathbf{6}\,(5).$

Consider the two primitive Pythagorean triangles $7^2+24^2=5^{2^2}$ and $119^2+120^2=13^{2^2}.$ In this case $t=2.$

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 $\frac{1}{2}\,(119)(120)=\mathbf{28}\,(255).$

Consider the two primitive Pythagorean triangles $164833^2+354144^2=5^{2^4}$ and $815616479^2+13651680^2=13^{2^4}.$ In this case $t=4.$

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 $2^t\left(2^{t+1}-1\right).$