Pythagorean Triples

If you've had high school geometry, you know about the Pythagorean Theorem. It states that the sum of the squares of the sides adjacent to the right angle in a right triangle is equal to the square of the side opposite the right angle, often written $a^2 + b^2 = c^2$. Pythagorean triples are solutions to this equation where $a$, $b$, and $c$ are all positive integers. We explicitly exclude solutions where $a$ or $b$ is zero. (For example, if $a = 0$, then $b = c$, and any positive integer will work for $b$. Not very interesting.) Suppose we want to find a method for determining all the solutions for a given $a$. So, for example, if $a = 27$, are there there any solutions where $b$ and $c$ are positive integers? If so, what are they? And can we be sure we've found all of them?

Let's begin by recasting the problem like this:

$$ \begin{align} c & = b + m \\ a^2 + b^2 & = (b + m)^2 \\ a^2 & = m(2b + m) \end{align} $$

Clearly $m$ must be a factor of $a^2$. One approach to solving the problem, therefore, would be to find all the factors of $a^2$ and see which (if any) lead to a solution. In our example, $27^2 = 729$ has factors $1, 3, 9, 27, 81, 243, $ and $729$. This leads to the following family of potential solutions:

$$ \begin{align} m &= 1:& 729 &= 1\cdot(2b + 1),& b &= 364,& c &= 365 \\ m &= 3:& 729 &= 3\cdot(2b + 3),& b &= 120,& c &= 123 \\ m &= 9:& 729 &= 9\cdot(2b + 9),& b &= 36,& c &= 45 \\ m &= 27:& 729 &= 27\cdot(2b + 27),& b &= 0,& c &= 27 \\ m &= 81:& 729 &= 81\cdot(2b + 81),& b &= -36,& c &= -45 \\ m &= 243:& 729 &= 243\cdot(2b + 243),& b &= -120,& c &= -123 \\ m &= 729:& 729 &= 729\cdot{2b + 729},& b &= -364,& c &= -365 \end{align} $$

Some of the potential solutions do not yield positive integers for $b$. In fact only the first three do. It looks as if we can ignore factors where $m \ge a$. Let's see if we can formalize this intuition.

First, if $m = a$, then clearly $b = 0$, so we can ignore factors where $m = a$. Second, since $a^2 = m(2b + m)$, we know that $m > a$ must mean $2b + m < a$. Therefore, $2b < a - m$ However, since $m > a$, $a - m < 0$, so $2b < 0$, which can only be true if $b < 0$. So if $m > a$, then $b < 0$. Since we are only interested in positive integer solutions, we can ignore factors $m$ of $a^2$ where $m \ge a$.

Let's try another example. Let's let $a = 105$. Then $a^2 = 11,025$, which has $27$ factors of which there are $13$ less than $105$: $1, 3, 5, 7, 9, 15, 21, 25, 35, 45, 49, 63, 75$. This leads to $13$ solutions:

$$ \begin{align} m &= 1:& 11,025 &= 1\cdot(2b + 1),& b &= 5,512,& c &= 5,513 \\ m &= 3:& 11,025 &= 3\cdot(2b + 3),& b &= 1,836,& c &= 1,839 \\ m &= 5:& 11,025 &= 5\cdot(2b + 5),& b &= 1,100,& c &= 1,105 \\ m &= 7:& 11,025 &= 7\cdot(2b + 7),& b &= 784,& c &= 791 \\ m &= 9:& 11,025 &= 9\cdot(2b + 9),& b &= 608,& c &= 617 \\ m &= 15:& 11,025 &= 15\cdot(2b + 15),& b &= 360,& c &= 375 \\ m &= 21:& 11,025 &= 21\cdot(2b + 21),& b &= 252,& c &= 273 \\ m &= 25:& 11,025 &= 25\cdot(2b + 25),& b &= 208,& c &= 233 \\ m &= 35:& 11,025 &= 35\cdot(2b + 35),& b &= 140,& c &= 175 \\ m &= 45:& 11,025 &= 45\cdot(2b + 45),& b &= 100,& c &= 145 \\ m &= 49:& 11,025 &= 49\cdot(2b + 49),& b &= 88,& c &= 137 \\ m &= 63:& 11,025 &= 63\cdot(2b + 63),& b &= 56,& c &= 119 \\ m &= 75:& 11,025 &= 75\cdot(2b + 75),& b &= 36,& c &= 111 \end{align} $$

That seemed to go pretty well. Let's make another trial. Lets choose $a = 10$, $a^2 = 100$. There are 9 factors, of which 4 are less than 10: 1, 2, 4, 5. Let's see what happens if we try them.

$$ \begin{align} m &= 1,& 100 &= 1\cdot(2b + 1),& b &= 49\frac 1 2,& c &= 50\frac 1 2 \\ m &= 2,& 100 &= 2\cdot(2b + 1),& b &= 24,& c &= 26 \\ m &= 4,& 100 &= 4\cdot(2b + 1),& b &= 10\frac 1 2,& c &= 14\frac 1 2 \\ m &= 5,& 100 &= 5\cdot(2b + 1),& b &= 7\frac 1 2,& c &= 12\frac 1 2 \end{align} $$

Only one of these solutions satisfies the requirement that all the numbers be positive integers. To see why, let's consider what happens if $a$ is even. If $a$ is even, then $m(2b + m)$ must also be even. However, whenever $m$ is odd, the expression $m(2b + m)$ must also be odd. Therefore $m$ odd can never lead to a solution for $a$ even; $m$ must also be even. What about the case above where $m = 4$? Let's consider the case where $a = {2^n}p$ and $p$ is odd. Then ${2^{2n}}p^2 = m(2b + m)$ Clearly, if $m = 2^{2n}$, then $p = 2b + 2^{2n}$. However, this contradicts the stipulation that $p$ must be odd. Similar reasoning applies if $p = qr$ and $m = 2^{2n}q$. Therefore, $m = 2^{2n}$ cannot lead to a solution. In summary, for $a$ even, there are no integer solutions when $m$ is odd or when $m$ contains all the $2$s in $a^2$.

Let's take a look at one more example. Suppose $a = 360$, $a^2 = 129,600$. There are $105$ factors of $129,600$ of which $52$ are less than $360$. We can count the factors by looking at the prime factorization of $129,600$. Let's start with the prime factorization of $360$.

$360 = 2^{3}\cdot3^{2}\cdot5^{1}$

Squaring it gives a prime factorization for $129,600$ of:

$129,600 = 2^{6}\cdot3^{4}\cdot5^{2}$

Counting $0$ as an exponent, we get $7 \cdot 5 \cdot 3 = 105$ total factors of $129,600$. Of those $5 \cdot 3 = 15$ are odd. Another $15$ contain the maximum power of $2$. Also just less than half, $52$, are less than $360$. Here is a table that shows which factors lead to solutions. All in all, there are $37$ values for $m$ that lead to a solution.

	2⁰	2¹	2²	2³	2⁴	2⁵	2⁶
3⁰ ∙ 5⁰	1	2	4	8	16	32	64
3¹ ∙ 5⁰	3	6	12	24	48	96	192
3² ∙ 5⁰	9	18	36	72	144	288	≥ 360
3³ ∙ 5⁰	27	54	108	216	≥ 360	≥ 360	≥ 360
3⁴ ∙ 5⁰	81	162	324	≥ 360	≥ 360	≥ 360	≥ 360
3⁰ ∙ 5¹	5	10	20	40	80	160	320
3¹ ∙ 5¹	15	30	60	120	240	≥ 360	≥ 360
3² ∙ 5¹	45	90	180	≥ 360	≥ 360	≥ 360	≥ 360
3³ ∙ 5¹	135	270	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360
3⁴ ∙ 5¹	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360
3⁰ ∙ 5²	25	50	100	200	≥ 360	≥ 360	≥ 360
3¹ ∙ 5²	75	150	300	≥ 360	≥ 360	≥ 360	≥ 360
3² ∙ 5²	225	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360
3³ ∙ 5²	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360
3⁴ ∙ 5²	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360	≥ 360

I leave it to the reader to find the solutions.

Ordering Chaos

Pythagorean Triples